Physics - Holding a Block Against a Wall

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This lecture will cover Newton's Second Law: F = ma.
Problem Text:
A block of 2.0kg is being held against a vertical wall by a 3.0N force directed at an angle of 15* above the horizontal. What coefficient of friction is required to keep the block from sliding.

[@miguel1502009]

i feel my left ear is a lot smarter now

[@i.cantdrinkmilk]

mlbra11 it works in both ears??

[@miguel1502009]

@@i.cantdrinkmilk it's only working on the left ear

[@fridamoreno9689]

I thought my AirPods were broken 😭😭

[@MichelvanBiezen]

Gareth, The force is directed at an angle of 15 degrees above the horizontal and that is how it is drawn as well. Remember that a vector can be moved to any location as long as the magnitude and direction doesn't change. So you can move the vector to the left so that the tail starts at the wall and then the vector will "appear" above the horizontal. It is actually the exact same situation with the exact same outcome.

[@garethm3171]

Still a tad confused by your explanation Michael, I agree vectors can be moved as long as their direction doesn't change; but in this case the question states that the vector is above the horizontal whereas in the video it's drawn below the horizontal with reference to the dotted line. If the vector is drawn above the horizontal dotted line it points downwards as you would expect it do (with the tip touching the block), then the Y component of that vector would be in the same direction as mg; however if it is drawn below the horizontal dotted line then it has to point upwards (again tip touching the block) as therefore the Y component points in the same direction as the friction force. I suppose in this case the orientation of the vector has changed from pointing down to pointing up. :-)

[@MichelvanBiezen]

Gareth M Gareth, Maybe this will help: Think about the sentence: "The force is directed at an angle of 15 degrees above the horizontal" Before you do anything else, think about the meaning of that sentence.

[@garethm3171]

I have Michel hence my observations above! Clearly it's a moot point as to what constitutes above or below a horizontal reference line! :-)

[@phynos8936]

Gareth M The force vector is aiming 15 degrees above the horizontal. Where the arrow is on the vector is key. In this diagram, If it were on the bottom-right, the force is directed away from the wall at 15 degrees below the horizontal. This force aims 15 degrees below the horizontal. Since it is on the top-left, it is a force directed towards the wall at 15 degrees above the horizontal. This force aims 15 degrees above the horizontal.

[@MichelvanBiezen]

Phynos Excellent explanation!

[@frostyshock643]

Actually 6.5 can be the answer as even though in most relevant applications µ

[@garethm3171]

In the question it says the force is directed at an angle of 15deg above the horizontal. However in your diagram you have drawn it at an angle of 15deg below the horizontal.

[@johndingo4633]

I know it was 2 years ago, but for anyone who is still wondering (like me). I think I got the solution: we must think the angle forming from the origin of the vector, not the end. If you look at minute 2:34 Michael draws a second 15° angle starting from a line parallel to the floor. It's like a mirrored unit circle. Try mirroring the whole problem.

[@joeyborja423]

To some who might still be reading this comment, here's my take. At first try, I also drew it with the tip on the block and the tail pointing upward to the right and drew an angle 15deg with horizontal line I drew from the tip of the arrow toward left, BUT, intuition told me that, 1) I had to solve the problem according to what the intention of the owner of the problem would be and not mine, 2) to support a block against a wall and to keep it from sliding down due to "mg", it makes no sense applying the force downwards, and 3) in, archery, when we shoot an arrow and "direct" it at a 15deg angle, we don't usually measure the angle from the tip of the arrow end but from the tail end (near our eyesight) with respect to our horizontal line of sight, which in that same sense, Mr Biezen drew the other 15deg angle at 2:36 with respect to the x-component supporting his point that from there the main force itself is directed 15deg above the horizontal up towards the block.

[@pirate1619]

@@johndingo4633 Thank You! 🥹❤❤

[@pirate1619]

@@joeyborja423 Thank You Too! ❤❤

[@daredevil-mf4zk]

Could you explain more as to why friction force is pointing up?

[@MichelvanBiezen]

The direction of the friction force will be opposite to the direction of motion if there was no friction.

[@williamwelmans8648]

A lovely Lesson! Thank you very much! But I get a coefficient of friction of 0.41 when a force of 300 Newtons is applied. I'm most probably wrong but can not fathom as to why.

[@brendanburvill6302]

Thanks mate. good to remember n is the direction perpendicular to the surface!

[@yim_e]

what force would be required for the block to start sliding upwards at a constant velocity?

[@oseamiya3663]

So the coefficient of friction is kinetic or static?

[@MichelvanBiezen]

If the block is not moving, then the coefficient of friction is static.

[@arieljackson9309]

These are so helpful! Great explanations :)

[@armygirl-30]

Sir what about the reaction force to fsintheta

[@MichelvanBiezen]

There is no reaction force to F sin(theta).

[@armygirl-30]

@@MichelvanBiezen Sir sorry for asking again but according to Newton's third law of motion there is always an equal and opposite reaction and as there is a reaction force to fcostheta in the opposite direction there should be a reaction force to fsintheta as well as they both are the components of the force which is being applied on the block Thanks in Anticipation 😊

[@xYondaimexJD]

why would coefficient of friction be dependent on applied force? shouldnt it be a constant?

[@MichelvanBiezen]

The coefficient of friction does not vary as a function of applied force.The problem is asking: what would the coefficient of friction have to be so that the block will not slide?

[@xYondaimexJD]

but the way the question was addressed makes it seem like mu was a function of applied force because you spent time comparing mu = 6.5 with 3.0 N applied force with that of mu = 0.41 with 30.0 N. I feel like the way the question was worded could potentially give the wrong idea to beginning students. I'm not targeting you or anything...but this is a problem with physics textbook questions as well.

[@MichelvanBiezen]

You are correct that it is often difficult to interpret word problems of any type, including those in physics. This question (in my opinion) appears clear about what they are asking, but what is clear to me does not necessarily appear clear to others. (The mystery of communication).

[@joshuaronisjr]

Why can't the force of static friction be more than 1? What if the brick was superglued to the wall or something, then wouldn't that force have to be much more than 1?

[@joshuaronisjr]

I mean coefficient, because regardless of how small the normal force is, the block will stay up

[@MichelvanBiezen]

There are unusual circumstances where the coefficient of friction can be greater than 1 as you pointed out.

[@MilesSpearman]

I was wondering if you knew how to accomplish this problem: Starting from rest, you are pushing a box (m=4kg) horizontally with a constant force F=30N. After pushing it for a distance d=5m, the box reaches a hill inclined at theta=25⁰. You stop pushing the box as soon as it reaches the hill. The coeficient of the kinetic friction between the biz and the road is =0.4 and the hill is frictionless. How far up the hill is the box going to go? I can email you what i did but i got an answer of 0.6787 meters and I dont think that is correct.

[@MichelvanBiezen]

W = F * d = 30N * 5 m = 150 J = KE of the box after pushing. This will be converted to PE = mgh + W friction = KE initial h = KE - W friction) / mg = ( 150 J - mg cos(theta) * mu) / (4 kg)*(9.8 m/sec^2) = Then d (up the ramp) = h/sin (theta) We have a number of examples in our playlists just like that.

[@mackenziemahdawi8872]

How would you find the velocity after it had traveled a distance?? I thought I could do ma=Fk+F-mg and solve for a and then V^2=V(0)^2+2aD and solve for V but my answer was wrong :( Do you have a Lecture Video for that?

[@MichelvanBiezen]

Take a look at this playlist: PHYSICS 2 MOTION IN ONE DIMENSION kzbin.info/aero/PLX2gX-ftPVXUlFGXu6QHg7dHowAw3zKoA or this one: PHYSICS 2.5 - 1D MOTION : GRAPHIC SOLUTIONS kzbin.info/aero/PLX2gX-ftPVXWyfZ4OqIP1aLKf_wxRb33Z

[@mackenziemahdawi8872]

Thank you Dr. Biezen

[@joshua7015]

work energy theorem

[@loganhonts5715]

what if the block is sliding down at a constant speed (but not accelerating)? what is the upward force required?

[@MichelvanBiezen]

The net force must then be zero and thus it will be the weight of the block minus the friction force.

[@jellyfrancis]

Why that the 0≤μ≤1 ?🤔❣️

[@MichelvanBiezen]

The coefficient of friction is derived experimentally. For just about every typical surface the coefficient of friction is measured to be between 0 and 1. If it were greater than 1 it would require more force to drag the object across the surface than lifting it in the air.

[@jellyfrancis]

@@MichelvanBiezen and that is possible isn't it 🤔💗

[@MichelvanBiezen]

For surfaces that are flat with only the roughness of the material contributing to the coefficient of friction you would not expect to find a coefficient of friction greater than 1

[@jellyfrancis]

@@MichelvanBiezen ok ❤️😊

[@tsehayenegash8394]

not hearable. have no sound

[@MichelvanBiezen]

Our old videos were recorded in mono sound (not stereo) so the sound will come out of one speaker.

[@MilesSpearman]

the reason the mu was so large was because the mass was so large, with such small force in the opposite direction, correct?

[@MichelvanBiezen]

That is correct.

[@TheSwagzmania]

What happens per say if the force acting on the block acts in the horizontal direction without an upward component. I.e the force was perpendicular to the block?

[@MichelvanBiezen]

That makes it easier as you don't have to consider the vertical component of the force. (The normal force then equals the F applied).

[@TheSwagzmania]

Okay. Cause I'm trying to understand a question like this. A body(2kg) is held against a wall by a horizontal force(60N) perpendicular to the body. Coefficient of static friction = 0.50.We were told the frictional force between the cube and the wall was 20N. I'd expect in an instance such as this, Fnet in the Y direction is 0 since the body is at rest, Fnet in the x direction would be 60-friction=0 since the body again is at rest. and thus 60=friction=normal(60N)*mew(s) or 60*0.5. Why then wouldn't the frictional force between the cube and wall be 30N?

[@MichelvanBiezen]

The friction force will be equal to the weight of the object, even though the normal force x coefficient is greater. The friction force will adjust to what it needs to be to stop the block from moving up to the maximum possible. (If the block had a mass of 4 kg the 60 N force would not be large enough to keep the block from sliding.)

[@karomatulloradzhabov8346]

Thanks a lot

[@anilkumar-ic5ni]

A cooler is standing on angle mounted on the wall Having weight 100 kg How much long it stand with The cement and wood are holding this cooler and heavy gaze iron t angle mounted on the wall Its own weight is 30 kg So how much weathering destroyed it In India we see this type of situation and many decade's are passed but we don't care about this One's installed Is their any study which show the weathering and weight proportion

[@MichelvanBiezen]

I don't know of any studies like that. But what manufacturers typically do, is test products for aging and wear in an accelerated manner so they don't have to wait years for the result. (It is exposed to corrosive materials, heat, cold, radiation, etc.)

[@nabarajbaral2063]

No work is done when a person is standing by carrying a load of 50kg for one hour.why?

[@TheRaviKr]

help ful video

[@kiruthika___attheglobalcit4134]

👍👍👍👍👍👍👍👍👍

[@antoniomantovani3147]

People says that tonno it's fish, and not meat. This Is silly