TURNING EFFECT OF A FORCE : LESSON 2

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Kind Tuition Academy

Kind Tuition Academy

Күн бұрын

Пікірлер: 120
@dengdee1484
@dengdee1484 2 жыл бұрын
U are the best teacher. Your Quotes of the lesson really instill a great impact in me bro.Thank God 🙏 for your presence.
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Wow, thanks friend 🤝. I'm humbled by your comments. I'm very happy to hear that you are benefiting from this channel, we thank God for everything. Be blessed too 🙏🙏🙏.
@dengdee1484
@dengdee1484 2 жыл бұрын
U are welcome
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
👍
@AJANG-MAYEN
@AJANG-MAYEN 3 жыл бұрын
Much appreciation to you mwalimu.You have actually taught this lesson well from the first lesson to the second lesson which is the last one 🙏🙏🙏
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Wow, great to hear this. Welcome bro. Thank you too for following my lessons.🤝
@AJANG-MAYEN
@AJANG-MAYEN 3 жыл бұрын
@@kindtuitionacademy welcome
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
👍
@Jackiembugua-w4x
@Jackiembugua-w4x Жыл бұрын
Thankyou for this lesson i now understand turning effect of a force
@kindtuitionacademy
@kindtuitionacademy Жыл бұрын
Welcome friend. I'm humbled and happy to hear that from you. Thanks for your comment, I wish you all the best in your studies friend 🙏
@celestineoindo9234
@celestineoindo9234 5 ай бұрын
Thanks for the lesson Turning effect is now very easy than i thought
@kindtuitionacademy
@kindtuitionacademy 5 ай бұрын
Wow, I'm Glad to hear that! Thanks for your comment friend 🙏.
@JabirAdesh
@JabirAdesh Жыл бұрын
Thank you teacher for helping me to understand this topic
@kindtuitionacademy
@kindtuitionacademy Жыл бұрын
Welcome friend. Thanks for your comment.
@elvisduke8403
@elvisduke8403 4 жыл бұрын
In calculating the moment of couple ...and if the forces are different which one would you choose
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
Good question. Couple forces are always equal in size/magnitude, oppositely directed and parallel to each other. So there's NO case where you will be asked to find the moment of couple of the forces which are different in size/magnitude because in that case the forces will not form a couple 😂😂😂. N/B For any two or more forces to form a couple, they forces must be EQUAL in magnitude, parallel and oppositely directed.👍
@josephowira8931
@josephowira8931 2 жыл бұрын
wow mwalimu thank you for the inncredible teaching
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Welcome friend 🤝. Thanks too for choosing to follow my lessons, I really appreciate 🙏.
@hindojama4487
@hindojama4487 3 жыл бұрын
A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by the two vertical spring balance c and d. Balance c is 20cm from the other end. Find the reading on each balance.
@fredkaranja4881
@fredkaranja4881 Жыл бұрын
Gud work sir keep going.
@kindtuitionacademy
@kindtuitionacademy Жыл бұрын
Thanks friend, I really appreciate 🙏.
@Horrific_lucious
@Horrific_lucious 4 ай бұрын
Thanks for helping me so much 🙏🏿
@kindtuitionacademy
@kindtuitionacademy 4 ай бұрын
Welcome friend. I'm humbled 🙏.
@omarsuleiman2735
@omarsuleiman2735 4 жыл бұрын
Why dont u continue please bro...ur helping many students please
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
Have no doubt I'll finish the whole syllabus upto form four work it's only that right now I'm somehow held-up with other things but I promise to continue whenever I get any free time. Thank you for your concern bro.🙏
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
Kindly share this channel to your colleagues. It's my wish for every student to benefit from this channel irregardless of their circumstances, I can't achieve this without your support 🙏. Thanks bro.
@physicsbrainik7364
@physicsbrainik7364 3 жыл бұрын
what is the difference between parallel and antiparallel forces 🙂🙂
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Parallel forces are parallel to each other (i.e they do not intersect) while antiparallel forces do not intersect (are parallel to each other) and strictly move in opposite directions, i.e if one force is moving towards the right hand direction, then the other force must move towards the left-hand direction. Thanks for your comment 🙏.
@salydianjeru8959
@salydianjeru8959 4 жыл бұрын
Would you be starting the lessons by calculating the last questions from the previous lesson
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
Yeah it's possible. Although I usually solve the exercise questions in the comment section. If you get a challenging question or a concept that you do not understand well, just drop a comment on that particular video and I'll explain to you how to handle it where possible. Thank you so much for your positive suggestion and for following my lessons, I really appreciate. Be blessed my siz🙏🙏🙏.
@ibrahimm66666
@ibrahimm66666 3 жыл бұрын
@@kindtuitionacademy how can I see answers of the questions you gave
@JamesMumbo-ye8tq
@JamesMumbo-ye8tq 5 ай бұрын
Tr please assist on how to do them because my answers are different
@lydiawanjiku3975
@lydiawanjiku3975 3 жыл бұрын
Thank you for the good job you are doing
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Wow, I'm humbled by your comment, it gives me the energy to keep producing more videos. Thanks alot Lydia🙏🙏🙏.
@officialsikdoc
@officialsikdoc 2 жыл бұрын
I highly appreciate your work but could you assist me with question 2 please
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Thanks friend 🤝. Ok. Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@irenengaruiya7384
@irenengaruiya7384 2 жыл бұрын
Well appreciated although could you kindly assist with the very last question(number 2)
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@gladboya2236
@gladboya2236 2 жыл бұрын
Please help in question 2 of the lastt exercise
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@ZakiCade-or8ej
@ZakiCade-or8ej 8 ай бұрын
How did you do that question 1
@vanessanyanchama9768
@vanessanyanchama9768 3 жыл бұрын
Thanks but the excersise you gave I have a challenge in the second one pliz do it
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Welcome friend. Here is the working for the questions in the exercise 👉👉👉 Question 1: Uniform half metre rule means the total length of the bar is (100cm*1/2 = 50cm or 0.5m). Therefore the distance between the couple forces ( i.e 20N to 20N) is : 50cm - ( 7.6cm + 12.4cm)= 30cm or 30/100= 0.3m. Moment of the couple= One of the forces* distance between the two forces forming the couple I.e 20N. Therefore: Moment of the couple of couple = F*d = 20N*0.3m = 6Nm 👉👉Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏
@vanessanyanchama9768
@vanessanyanchama9768 3 жыл бұрын
Thanks well understood
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Welcome friend 🙏.
@mbodzejeffa
@mbodzejeffa 6 ай бұрын
What is the difference between parallel and anti parallel force s
@kindtuitionacademy
@kindtuitionacademy 6 ай бұрын
Parallel forces (are forces which are parallel to each other i.e they can never intersect each other and move in the same ⬇️⬇️) while anti-parallel forces (are parallel forces i.e they can never intersect each other but are moving in opposite directions ⬇️⬆️). I hope you're able to differentiate the two. Thanks for your comment 🙏.
@anisaduane4162
@anisaduane4162 3 жыл бұрын
Can you please do a vedio of hookes law and pressure when calculating using a u tube manometer
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
I'm working on Hooke's law. I'll be apploading it probably in early March. About calculations on U-Tube manometer I've already covered that in a video called "Pressure: Lesson 7" check it here👉👉👉kzbin.info/www/bejne/oKa2YXmOecaGsJo
@Jackiembugua-w4x
@Jackiembugua-w4x Жыл бұрын
Can you show me how to do number 2 in the exercise section
@kindtuitionacademy
@kindtuitionacademy Жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@jerokhate1040
@jerokhate1040 3 жыл бұрын
thank you, sir. more videos please
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Welcome Jero Khate. Thanks too for your comment . I'm working on that, Sound : Lesson 3 is dropping tomorrow, stay tuned🙏.
@maryakoth2879
@maryakoth2879 6 ай бұрын
Mwalimu that number 2 of the exercise
@abdullahiibrahim9240
@abdullahiibrahim9240 Күн бұрын
please teacher tackle excercise number 2
@vanessanyanchama9768
@vanessanyanchama9768 3 жыл бұрын
Teacher can you post more examples on the calculations on the topic waves
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Yes, I haven't completed the topic waves (1), so expect more examples in my next video which will be Waves (1) : Lesson 3. Thanks for your comment 🙏.
@rajnimalhotra8924
@rajnimalhotra8924 2 жыл бұрын
Thank you immensely
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Welcome friend 🤝. Thanks too for always supporting this channel, I really appreciate 🙏.
@trevasdat2993
@trevasdat2993 2 жыл бұрын
Help me on no2 parallel ex
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Ok Trevas. Here's the working 👇 Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Question 1: Uniform half metre rule means the total length of the bar is (100cm*1/2 = 50cm or 0.5m). Therefore the distance between the couple forces ( i.e 20N to 20N) is : 50cm - ( 7.6cm + 12.4cm)= 30cm or 30/100= 0.3m. Moment of the couple= One of the forces* distance between the two forces forming the couple I.e 20N. Therefore: Moment of the couple of couple = F*d = 20N*0.3m = 6Nm
@sumayayussuf7840
@sumayayussuf7840 3 жыл бұрын
I have calculated the exercise u have given but didn't get answer as your
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Ok, Yussuf, Maybe you went wrong somewhere. Here is the correct working 👉👉👉 Question 1: Uniform half metre rule means the total length of the bar is (100cm*1/2 = 50cm or 0.5m). Therefore the distance between the couple forces ( i.e 20N to 20N) is : 50cm - ( 7.6cm + 12.4cm)= 30cm or 30/100= 0.3m. Moment of the couple= One of the forces* distance between the two forces forming the couple I.e 20N. Therefore: Moment of the couple of couple = F*d = 20N*0.3m = 6Nm Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏
@abdirizaqdayib1231
@abdirizaqdayib1231 4 жыл бұрын
What do you mean tangitelly
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
Tangentially means the forces are acting through a tangent from the steering wheel. A tangent is a straight line or a plane that touches a curve or a curved surface at a single point but doesn't intersect that curve or curved surface. You will understand more about tangent lines and normal lines when you cover a form four topic in mathematics called "Linear motion/ Kinematics/Calculus". 🙏
@Juujuule1
@Juujuule1 2 жыл бұрын
Thank you so much teacher 🙂😄😄😄😄🤎🤎🤎🤎🤎
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Welcome 🤝. Thanks too for following my lessons 🙏.
@dieudonekabika1834
@dieudonekabika1834 2 жыл бұрын
Why are you denoting the perpendicular distance as d1.2? The I have been confused
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
d1.2N simply means " the perpendicular distance of the force (1.2N) from the pivot. Thanks for your comment 🙏.
@dieudonekabika1834
@dieudonekabika1834 2 жыл бұрын
Thank you sir. And what if you are told to calculate the tension in a string and also the reaction forces. What do we do?
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
It depends with the direction in which that tension force is acting in relation to the pivot i.e is the tension causing the bar to rotate in the clockwise or anti-clockwise direction. In most cases, reaction force will always act upwards . Thanks for your comment 🙏.
@dieudonekabika1834
@dieudonekabika1834 2 жыл бұрын
Thank you sir. Your work is highly appreciated. May God bless you
@ttcomedian5631
@ttcomedian5631 Жыл бұрын
Motion under gravity
@PrudenceMathu-kz7vd
@PrudenceMathu-kz7vd 6 ай бұрын
I cañ't get the answer of the question number2(last question)
@vashitwabera8758
@vashitwabera8758 2 жыл бұрын
Do question 2 please do question it o question too because it is too difficult for me to answerr it is too difficult for me to
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Ok. Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for always supporting this channel, I really appreciate 🙏🙏 🙏.
@halimaabdullahi8791
@halimaabdullahi8791 4 жыл бұрын
Can you please help me calculate no 2
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏
@aminambarak1342
@aminambarak1342 3 жыл бұрын
I heard that couple forces is out of the syllabus
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
I'm not sure, unless I confirm.
@elizabethmukonyo5827
@elizabethmukonyo5827 2 жыл бұрын
Please do number 2
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@AnnMuthoni-j5c
@AnnMuthoni-j5c Жыл бұрын
Thank you sir
@kindtuitionacademy
@kindtuitionacademy Жыл бұрын
Welcome friend. I'm humbled. Thanks too for always supporting this channel, I really appreciate 🙏.
@elvisduke8403
@elvisduke8403 4 жыл бұрын
Can you help me in calculating no 1
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
Question 1: Uniform half metre rule means the total length of the bar is (100cm*1/2 = 50cm or 0.5m). Therefore the distance between the couple forces ( i.e 20N to 20N) is : 50cm - ( 7.6cm + 12.4cm)= 30cm or 30/100= 0.3m. Moment of the couple= One of the forces* distance between the two forces forming the couple I.e 20N. Therefore: Moment of the couple of couple = F*d = 20N*0.3m = 6Nm
@abdimalikabdullahi8283
@abdimalikabdullahi8283 2 жыл бұрын
Please do number 2 exercise
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons, I really appreciate 🙏
@emilykageha7599
@emilykageha7599 3 жыл бұрын
Hi mwalimu..couple forces is the same as parallel or anti parallel forces? Please explain
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Hi too 🤝. Couple forces are usually anti-parallel and have the same magnitude/size. The term "Anti-parallel" in this case means that the two forces are parallel to each other ( i.e the forces can never intersect/meet each other) but are moving in opposite directions (i.e if one force is moving towards the right hand direction, then the other force must be moving towards the left hand direction). The term "same Magnitude/same size" in this case means that if one of the forces forming the couple is say, 5N then, the other force forming the couple must be 5N. Thanks for your comment 🙏.
@mbodzejeffa
@mbodzejeffa 6 ай бұрын
What about parallel forces
@omarsuleiman2735
@omarsuleiman2735 4 жыл бұрын
Wher is measurement 2 form 2 please bro
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
It's there immediately after" Magnetism Lesson 4". Just check well you'll see it. I've arranged those topics systematically.
@kindtuitionacademy
@kindtuitionacademy 4 жыл бұрын
Kindly share this channel with your colleagues. It's my wish for every student to benefit from this channel irregardless of their circumstances. I can't achieve this without your support 🙏. Thanks bro.
@Bar.b.i.e
@Bar.b.i.e 2 жыл бұрын
Ok.....u lost me somewhere....does that mean that parallel forces and anti parallel forces are both called couple forces?
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
No, parallel forces could be moving in the same direction hence they may not necessarily be couple forces (two parallel forces can only form couple forces if they have the same magnitude and they're moving in opposite direction) - anti-parellel forces may or may not form couple forces since they could be having different magnitudes/size (anti-parellel=parallel forces moving in opposite directions) N/B: For any pair of forces to form a couple, they MUST satisfy all the three conditions i.e: (1) must be parallel (2) must be moving in opposite directions (3) must have the same magnitude Thanks for your comment 🙏.
@JocaihChemongin
@JocaihChemongin 5 ай бұрын
Hy tr.is it true that parallel forces is not part of the syllabus or.....
@kindtuitionacademy
@kindtuitionacademy 5 ай бұрын
Yes, it's true. Thanks for your comment.
@JocaihChemongin
@JocaihChemongin 5 ай бұрын
Thanks 👍 tr,,,,,,,,.but even though it's not part of the topic at least i have the knowledge ,,,,, thanks once again ,,,you teach really well and you've really helped me through this holiday,,,,God bless you
@kindtuitionacademy
@kindtuitionacademy 5 ай бұрын
Welcome friend. I'm humbled friend. Be blessed too 🙏.
@KheirNisaa
@KheirNisaa Жыл бұрын
0:00 - 0:20
@josephowira8931
@josephowira8931 2 жыл бұрын
sorry but iwas asking for guidance in question 2 in turning effect of force lesson2
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Question 1: Uniform half metre rule means the total length of the bar is (100cm*1/2 = 50cm or 0.5m). Therefore the distance between the couple forces ( i.e 20N to 20N) is : 50cm - ( 7.6cm + 12.4cm)= 30cm or 30/100= 0.3m. Moment of the couple= One of the forces* distance between the two forces forming the couple I.e 20N. Therefore: Moment of the couple of couple = F*d = 20N*0.3m = 6Nm Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏
@maryanibrahimow9790
@maryanibrahimow9790 3 жыл бұрын
Sir Help me in no 2 .may u be blessed coz u have help us so much
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Welcome siz. I'm humbled by your comment. Be blessed too 🙏.
@maryanibrahimow9790
@maryanibrahimow9790 3 жыл бұрын
@@kindtuitionacademy thanks sir
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Welcome siz🙏.
@vanessanyanchama9768
@vanessanyanchama9768 3 жыл бұрын
More examples o the questions
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Have you attempted the exercise that I gave at the end of this lesson?
@ReshaSisi
@ReshaSisi 5 ай бұрын
Am a student and question 2 has become challenging please help me
@kindtuitionacademy
@kindtuitionacademy 5 ай бұрын
Ok, Maybe you went wrong somewhere. Here is the correct working 👉👉👉 Question 1: Uniform half metre rule means the total length of the bar is (100cm*1/2 = 50cm or 0.5m). Therefore the distance between the couple forces ( i.e 20N to 20N) is : 50cm - ( 7.6cm + 12.4cm)= 30cm or 30/100= 0.3m. Moment of the couple= One of the forces* distance between the two forces forming the couple I.e 20N. Therefore: Moment of the couple of couple = F*d = 20N*0.3m = 6Nm Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏
@tesloachchop7250
@tesloachchop7250 2 жыл бұрын
Number 2 please 🙏
@kindtuitionacademy
@kindtuitionacademy 2 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏.
@mjtj3652
@mjtj3652 4 жыл бұрын
Please help me in number 2
@kindtuitionacademy
@kindtuitionacademy 3 жыл бұрын
Questions 2: Converting all lengths/distances into SI units we get: 1N weight is placed at at 50cm mark = 50÷100 = 0.5 metre mark; 0.12N force is placed at 100cm mark = 100÷100 = 1metre mark. 0.18N force is placed at the 0cm mark = 0÷100 = 0 metre mark. Let the pivot be placed at the X metres mark. Since weight 0.18N is greater than weight 0.12N, it means that the pivot must be placed somewhere between weights 0.18N and 1N inorder for the metre rule to balance. If the pivot is between weights 0.18N and 1N, then weights/forces 1N and 0.12N will be on the right hand side of the pivot hence they will contribute to clockwise moments. Conversely, 0.18N weight/force will be on the left hand side of the pivot hence it will contribute to anticlockwise moment. From the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments measured from the pivot, i.e F * d + F * d = F * d 0.12N * (1m - X) + 1N * (0.5m - X) = 0.18N * (X - 0) On opening the brackets we get; 0.12 - 0.12X + 0.5 - X = 0.18X On collecting the like terms we get; 0.12 + 0.5 = 0.18X + X + 0.12X 0.62 = 1.3X Dividing both sides by 1.3 we get X = 0.62 ÷ 1.3 X = 0.4769 metres ( correct to 4 decimal places) X = 0.4769 * 100 = 47.69 Cm Since 0.18N is placed at 0Cm mark, the distance of the pivot from 0.18N weight will be 47.69Cm - 0Cm = 47.69Cm Thanks for following my lessons 🙏
@GameSphere05
@GameSphere05 8 ай бұрын
Please techer help me no 2
@GameSphere05
@GameSphere05 8 ай бұрын
Please teacher help me no 2
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