U are the best teacher. Your Quotes of the lesson really instill a great impact in me bro.Thank God 🙏 for your presence.
@kindtuitionacademy2 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 🙏🙏🙏.
@dengdee14842 жыл бұрын
U are welcome
@kindtuitionacademy2 жыл бұрын
👍
@AJANG-MAYEN3 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 🙏🙏🙏
@kindtuitionacademy3 жыл бұрын
Wow, great to hear this. Welcome bro. Thank you too for following my lessons.🤝
@AJANG-MAYEN3 жыл бұрын
@@kindtuitionacademy welcome
@kindtuitionacademy3 жыл бұрын
👍
@Jackiembugua-w4x Жыл бұрын
Thankyou for this lesson i now understand turning effect of a force
@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 🙏
@celestineoindo92345 ай бұрын
Thanks for the lesson Turning effect is now very easy than i thought
@kindtuitionacademy5 ай бұрын
Wow, I'm Glad to hear that! Thanks for your comment friend 🙏.
@JabirAdesh Жыл бұрын
Thank you teacher for helping me to understand this topic
@kindtuitionacademy Жыл бұрын
Welcome friend. Thanks for your comment.
@elvisduke84034 жыл бұрын
In calculating the moment of couple ...and if the forces are different which one would you choose
@kindtuitionacademy4 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.👍
@josephowira89312 жыл бұрын
wow mwalimu thank you for the inncredible teaching
@kindtuitionacademy2 жыл бұрын
Welcome friend 🤝. Thanks too for choosing to follow my lessons, I really appreciate 🙏.
@hindojama44873 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 Жыл бұрын
Gud work sir keep going.
@kindtuitionacademy Жыл бұрын
Thanks friend, I really appreciate 🙏.
@Horrific_lucious4 ай бұрын
Thanks for helping me so much 🙏🏿
@kindtuitionacademy4 ай бұрын
Welcome friend. I'm humbled 🙏.
@omarsuleiman27354 жыл бұрын
Why dont u continue please bro...ur helping many students please
@kindtuitionacademy4 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.🙏
@kindtuitionacademy4 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.
@physicsbrainik73643 жыл бұрын
what is the difference between parallel and antiparallel forces 🙂🙂
@kindtuitionacademy3 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 🙏.
@salydianjeru89594 жыл бұрын
Would you be starting the lessons by calculating the last questions from the previous lesson
@kindtuitionacademy4 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🙏🙏🙏.
@ibrahimm666663 жыл бұрын
@@kindtuitionacademy how can I see answers of the questions you gave
@JamesMumbo-ye8tq5 ай бұрын
Tr please assist on how to do them because my answers are different
@lydiawanjiku39753 жыл бұрын
Thank you for the good job you are doing
@kindtuitionacademy3 жыл бұрын
Wow, I'm humbled by your comment, it gives me the energy to keep producing more videos. Thanks alot Lydia🙏🙏🙏.
@officialsikdoc2 жыл бұрын
I highly appreciate your work but could you assist me with question 2 please
@kindtuitionacademy2 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 🙏.
@irenengaruiya73842 жыл бұрын
Well appreciated although could you kindly assist with the very last question(number 2)
@kindtuitionacademy2 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 🙏.
@gladboya22362 жыл бұрын
Please help in question 2 of the lastt exercise
@kindtuitionacademy2 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-or8ej8 ай бұрын
How did you do that question 1
@vanessanyanchama97683 жыл бұрын
Thanks but the excersise you gave I have a challenge in the second one pliz do it
@kindtuitionacademy3 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 🙏
@vanessanyanchama97683 жыл бұрын
Thanks well understood
@kindtuitionacademy3 жыл бұрын
Welcome friend 🙏.
@mbodzejeffa6 ай бұрын
What is the difference between parallel and anti parallel force s
@kindtuitionacademy6 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 🙏.
@anisaduane41623 жыл бұрын
Can you please do a vedio of hookes law and pressure when calculating using a u tube manometer
@kindtuitionacademy3 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 Жыл бұрын
Can you show me how to do number 2 in the exercise section
@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 🙏.
@jerokhate10403 жыл бұрын
thank you, sir. more videos please
@kindtuitionacademy3 жыл бұрын
Welcome Jero Khate. Thanks too for your comment . I'm working on that, Sound : Lesson 3 is dropping tomorrow, stay tuned🙏.
@maryakoth28796 ай бұрын
Mwalimu that number 2 of the exercise
@abdullahiibrahim9240Күн бұрын
please teacher tackle excercise number 2
@vanessanyanchama97683 жыл бұрын
Teacher can you post more examples on the calculations on the topic waves
@kindtuitionacademy3 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 🙏.
@rajnimalhotra89242 жыл бұрын
Thank you immensely
@kindtuitionacademy2 жыл бұрын
Welcome friend 🤝. Thanks too for always supporting this channel, I really appreciate 🙏.
@trevasdat29932 жыл бұрын
Help me on no2 parallel ex
@kindtuitionacademy2 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 🙏.
@kindtuitionacademy2 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
@sumayayussuf78403 жыл бұрын
I have calculated the exercise u have given but didn't get answer as your
@kindtuitionacademy3 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 🙏
@abdirizaqdayib12314 жыл бұрын
What do you mean tangitelly
@kindtuitionacademy4 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". 🙏
@Juujuule12 жыл бұрын
Thank you so much teacher 🙂😄😄😄😄🤎🤎🤎🤎🤎
@kindtuitionacademy2 жыл бұрын
Welcome 🤝. Thanks too for following my lessons 🙏.
@dieudonekabika18342 жыл бұрын
Why are you denoting the perpendicular distance as d1.2? The I have been confused
@kindtuitionacademy2 жыл бұрын
d1.2N simply means " the perpendicular distance of the force (1.2N) from the pivot. Thanks for your comment 🙏.
@dieudonekabika18342 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?
@kindtuitionacademy2 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 🙏.
@dieudonekabika18342 жыл бұрын
Thank you sir. Your work is highly appreciated. May God bless you
@ttcomedian5631 Жыл бұрын
Motion under gravity
@PrudenceMathu-kz7vd6 ай бұрын
I cañ't get the answer of the question number2(last question)
@vashitwabera87582 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
@kindtuitionacademy2 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 🙏🙏 🙏.
@halimaabdullahi87914 жыл бұрын
Can you please help me calculate no 2
@kindtuitionacademy3 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 🙏
@aminambarak13423 жыл бұрын
I heard that couple forces is out of the syllabus
@kindtuitionacademy3 жыл бұрын
I'm not sure, unless I confirm.
@elizabethmukonyo58272 жыл бұрын
Please do number 2
@kindtuitionacademy2 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 Жыл бұрын
Thank you sir
@kindtuitionacademy Жыл бұрын
Welcome friend. I'm humbled. Thanks too for always supporting this channel, I really appreciate 🙏.
@elvisduke84034 жыл бұрын
Can you help me in calculating no 1
@kindtuitionacademy4 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
@abdimalikabdullahi82832 жыл бұрын
Please do number 2 exercise
@kindtuitionacademy2 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 🙏
@emilykageha75993 жыл бұрын
Hi mwalimu..couple forces is the same as parallel or anti parallel forces? Please explain
@kindtuitionacademy3 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 🙏.
@mbodzejeffa6 ай бұрын
What about parallel forces
@omarsuleiman27354 жыл бұрын
Wher is measurement 2 form 2 please bro
@kindtuitionacademy4 жыл бұрын
It's there immediately after" Magnetism Lesson 4". Just check well you'll see it. I've arranged those topics systematically.
@kindtuitionacademy4 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.e2 жыл бұрын
Ok.....u lost me somewhere....does that mean that parallel forces and anti parallel forces are both called couple forces?
@kindtuitionacademy2 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 🙏.
@JocaihChemongin5 ай бұрын
Hy tr.is it true that parallel forces is not part of the syllabus or.....
@kindtuitionacademy5 ай бұрын
Yes, it's true. Thanks for your comment.
@JocaihChemongin5 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
@kindtuitionacademy5 ай бұрын
Welcome friend. I'm humbled friend. Be blessed too 🙏.
@KheirNisaa Жыл бұрын
0:00 - 0:20
@josephowira89312 жыл бұрын
sorry but iwas asking for guidance in question 2 in turning effect of force lesson2
@kindtuitionacademy2 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 🙏.
@kindtuitionacademy2 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 🙏
@maryanibrahimow97903 жыл бұрын
Sir Help me in no 2 .may u be blessed coz u have help us so much
@kindtuitionacademy3 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 🙏
@kindtuitionacademy3 жыл бұрын
Welcome siz. I'm humbled by your comment. Be blessed too 🙏.
@maryanibrahimow97903 жыл бұрын
@@kindtuitionacademy thanks sir
@kindtuitionacademy3 жыл бұрын
Welcome siz🙏.
@vanessanyanchama97683 жыл бұрын
More examples o the questions
@kindtuitionacademy3 жыл бұрын
Have you attempted the exercise that I gave at the end of this lesson?
@ReshaSisi5 ай бұрын
Am a student and question 2 has become challenging please help me
@kindtuitionacademy5 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 🙏
@tesloachchop72502 жыл бұрын
Number 2 please 🙏
@kindtuitionacademy2 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 🙏.
@mjtj36524 жыл бұрын
Please help me in number 2
@kindtuitionacademy3 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 🙏