# How to get the force by Block A?

• zerogoal
In summary, the conversation discusses how to draw a free-body diagram of block A while it is moving with block B as a unit. The correct orientation of the force vectors is important, with the normal force and weight arrows being equal in length and the friction arrow being shorter. The direction of the friction force is also discussed, with one person suggesting it should be to the left and another suggesting it should be to the right. Eventually, it is agreed that friction should be in the direction of the displacement of block A.
zerogoal
A person pulls horizontally on block B in the figure (Intro 1 figure) , causing both blocks to move together as a unit.

While this system is moving, make a carefully labeled free-body diagram of block A if the table is frictionless.
Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.How do i do this?
Given four arrows: normal force, weight, friction and unlabeled vector

## The Attempt at a Solution

My first attempt was the normal force arrow pointing upwards from block A, friction and weight arrows pointing downwards from block A. Then the unlabeled vector arrow pointing to the right.

Then for my second attempt, i took away the friction arrow.But i still can't get the answer! Anyone can help me?? very urgent! left with one last try.

Last edited:
You need the friction arrow. Imagine if there was no friction between the two blocks, B would slide out from A and A would stay in the exact same place. So friction must be keeping A on top of B.

I input the friction arrow for my first attempt, but i did not get the correct ans.
May i know how long should my arrows be? Cause for my first attempt, my normal force arrow was the longest followed by the weight arrow then the friction arrow. Do i need to include the unlabeled vector as well?

The normal force should be equal to the weight - it's not moving in the y-direction, which means that the net force on the y-axis must = 0.

So that means the length of the normal and weight arrows should be equal right. and also the length of the friction arrow should be shorter than these 2 arrows?
Then do i still need to put the unlabeled vector arrow acting towards the right? Since Block B is the one that is being pulled, then i don't think i need to include the unlabeled vector arrow right? do correct me if I'm wrong. Thanks

So you don't have a set number of vectors?

I really don't think you need that unlabeled one - it doesn't exist a force that way, since friction is actually what makes the block also move forward.

But I'd think again about the direction of the friction force.

Nope i don't have the set of vectors. They only state unlabeled vector.
Shouldn't the friction force be acting downwards since Newton's 3rd law states that block A will exert a force equal in magnitude and opposite in direction with respect to block B?

No, think about what friction is first. Friction occurs when there's movement (or rather, an object is affected by friction when it's moving). If you push a block or something, it will stop, right? So, friction slows down velocity. But: velocity is parallel to displacement - and you know that they both are horizontal. If friction was a vertical force (which you say) it'd be 90 degrees to the velocity/displacement and would therefore not do any work on the object, i.e. not change the kinetic energy, and since kinetic energy is ½mv2, that means no change in velocity.

Am I making sense? I don't know what you know about mechanical energy and such, but yeah.

Yup I understood what you're saying.

Since force acting on Block B is towards the right, then the friction acting on Block A should be towards the left as these 2 forces should be in opposite direction. Am i right?

Well:

The displacement of Block B is to the right - friction between B and the floor is to the left.
Friction is what makes Block A move--shouldn't friction be in the direction of the displacement?

Block A wants to stay where it is, and imagine that A is fixed to a point and therefore can't move. So, B moves out from underneath A, but A is still wants to hold on to B. This is sort of the same scenario if you have B fixed at a point and move block A to the right--friction is opposite to A's displacement.. therefore to the right. (Hah, this is the weirdest explanation, but it's how I understand it; but when I think about it I use my hands and move them around and stuff.. ah well.)

I reckon the friction of block A is to the right, in this case. It makes more sense to me. Do you know where you draw friction vectors? I've been thaught that friction vectors go along the surface. (like, between A&B here.)

Sorry I'm a bit confused. I agree with you saying as block B moves to the right, the friciton would between block B and floor would be towards the left. However as block B moves to the right, friction would occur between surfaces of block A and B, but why is the friction acting to the right? Since the surface of block B is acting to the right, then shouldn't the friction be acting to the left? Sorry can help me clear my doubts? thanks a lot.

Anyway i can't draw the friction arrows along surface A and B as the question states that i must draw all forces from the centre of block A. How long should my friction arrow be? Shorter than normal vector arrow or the same length as the pull force?

I'm not going to say I'm the best with friction vectors (meaning, I've got some difficulties seeing their direction in front ot me), but I see it like this: Block A has three forces acting on it; normal force, weight and friction. Normal force & weight are in the y-direction and they cancel, so they don't matter for the displacement of the Block. However -- the displacement is to the right which means that there has to be an unbalanced force directed to the right in order for there to be displacement at all. Friction is the only one that's left.

Okay, then it doesn't matter about the friction vector. It was the type of thing that was important for my teacher in my previous Phyiscs course.

Edit: Okay, so I had a very brief discussion with a friend of mine (we're a bit preoccupied, matrix test tomorrow :|) and she said this:

"Because: friction always opposes the direction of displacement. The friction force felt on the bottom block is therefore to the left. Acc to Newtons third. opposite forces, makes the friction on the top block to the right."

Oh i see. Finally got what you meant. Thanks for clearing my doubts. Really appreciate your effort in spending time to help me solve this problem. Thanks a lot

## 1. How does the mass of Block A affect the force?

The force acting on an object is directly proportional to its mass. This means that as the mass of Block A increases, the force required to move it also increases.

## 2. What is the equation for calculating the force on Block A?

The equation for calculating force is F = ma, where F is the force, m is the mass, and a is the acceleration. This means that the force on Block A can be calculated by multiplying its mass by its acceleration.

## 3. How does the surface of Block A affect the force?

The surface of Block A can affect the force by providing friction. If the surface is rough, there will be more friction and thus more force required to move the block. If the surface is smooth, there will be less friction and less force required.

## 4. What is the difference between force and net force?

Force refers to any push or pull on an object, while net force specifically refers to the overall force acting on an object after all individual forces have been taken into account. In the case of Block A, the net force would be the sum of all forces acting on the block.

## 5. How can I increase the force on Block A?

The force on Block A can be increased by either increasing its mass or its acceleration. This can be achieved by adding more weight to the block or by applying a greater amount of force to move it. The surface on which the block is placed can also affect the force required to move it.

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