1) The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F causes both blocks to cross a distance of 5.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

http://session.masteringphysics.com/problemAsset/1000635/6/knight_Figure_08_32.jpg

Let A=top block, B=bottom block

There are 4 forces acting on the block A: Gravity, Normal force (B on A), F (the applied force), and static friction (B on A)

There are 5 forces acting on the block B: Grviaty, Normal force (earth on B), normal force (A on B), kinetic friction (earth on B), and static friction (A on B)

Fnet on B=static friction (A on B)-kinetic friction (earth on B)=(mass of B)(acceleration) and then find time using kinematics. Using this method, I got an answer of 1.75s and this is correct according to my textbook.

But this is not making sense to me. If the system is accelerating, then block A must be accelerating, i.e. net force on A should be to the right, and this means the applied force F is greater than the static friction of B on A. But if the applied force is greater than static friction, kinetic friction applies instead of static friction, right? If so, then block A is sliding on block B and I am running into terrible trouble on making sense of this problem...


Can someone please explain and help me making sense of this scenario? Thank you!

2) A house painter uses the chair and pulley arrangement of the figure to lift himself up the side of a house. The painter's mass is 70 kg and the chair's mass is 10 kg. With what force must he pull down on the rope in order to accelerate upward at 0.20m/s^2 ?
http://session.masteringphysics.com/problemAsset/1000645/5/knight_Figure_08_42.jpg

I am having some problems with this too, conceptually.

The answer is that if the painter pulls at a 400N force, he can accelerate upward at 0.20m/s^2.

Now, if, instead of pulling the rope, I hang a 400N block on the other end, WHY can't the painter accelerate upward at 0.20m/s^2? This is what I don't understand.

I hope someone with a good grasp at physics concepts can help me! Thanks!

Question 1:

But if the applied force is greater than static friction, kinetic friction applies instead of static friction, right?

Wrong.

Kinetic friction applies when the surfaces in question are moving relative to one another. When they are relatively stationary, static friction applies. So, if A is not slipping on B, static friction applies between A and B.

Question 2:

Why wouldn't hanging a 400N block on the other end work just as well as pulling the rope?

Question 1:



Wrong.

Kinetic friction applies when the surfaces in question are moving relative to one another. When they are relatively stationary, static friction applies. So, if A is not slipping on B, static friction applies between A and B.

Question 2:

Why wouldn't hanging a 400N block on the other end work just as well as pulling the rope?
Hi,

1) But fpr the top block, if the applied force is greater than the static friction, then the net force on it is to the right, so its will accelerate to the right. Then, since the net force is not zero, the applied force completely overcome the static friction and the top block will move...why isn't it kinetic friction when it's already moving?

2) I am not too sure about this...why?

2a) What forces are acting on the man in each case?
2b) Do you mean a block that weighs 400N in normal gravity? What would it weigh when accelerating downward at 2m/s?

Now, if, instead of pulling the rope, I hang a 400N block on the other end, WHY can't the painter accelerate upward at 0.20m/s^2? This is what I don't understand.

Here's the answer. Apply Newton's second law, F=ma. In the situation of him pulling his own rope, "m" is the mass of the painter (rope is assumed massless). When you put on a 400N block, the total mass of the system then becomes mass of painter + 40.8Kg. Since the total mass is now greater, the acceleration is less.

But if the applied force is greater than static friction, kinetic friction applies instead of static friction, right?
You have to first figure out if the force breaks the static friction to find out if they are sliding relative to one another. So to find the least time, you have to find the maximum force you can apply (i.e. F <= static friction). But, note that you will have to add the kinetic friction on the floor to your applied force, and have that be less than the static friction.

In other words, Fs - Fk >= Fapplied.

I think that's how you do it. I just did it really quickly, and I may have made a dumb mistake.

2a) What forces are acting on the man in each case?
2b) Do you mean a block that weighs 400N in normal gravity? What would it weigh when accelerating downward at 2m/s?

2a) Here, the man's hand is touching the rope, which complicates things...Will there be double tension? If so, does it mean the hand touching the rope actually helps him accelerate faster?

2b) Yes, I mean a block on earth weighing 400N.
It would weigh the same when accelerating downward at 2m/s, because force of gravity never changes (provided g is constant)

Kingwinner, for the first one you are free to consider the two blocks as a unit and therefore overlook internal forces since they both have the same acceleration when they are not slipping relative to one another. The tension force is thus exerted on an equivalent mass of 7kg. This will help you figure out what the acceleration of each of the two blocks is during the motion.

I might be wrong though. I don't have the best physics track record on this forum, lemme tell ya.

You have to first figure out if the force breaks the static friction to find out if they are sliding relative to one another. So to find the least time, you have to find the maximum force you can apply (i.e. F <= static friction). But, note that you will have to add the kinetic friction on the floor to your applied force, and have that be less than the static friction.

In other words, Fs - Fk >= Fapplied.

I think that's how you do it. I just did it really quickly, and I may have made a dumb mistake.

But you don't know the magnitude of the applied force so you can't used it to figure out if the top block is sliding on the lower block...

How can the top block accelerate (net force to the right) while we can still consider STATIC friction? (not kinetic friction)

Kingwinner, for the first one you are free to consider the two blocks as a unit and therefore overlook internal forces since they both have the same acceleration when they are not slipping relative to one another. The tension force is thus exerted on an equivalent mass of 7kg. This will help you figure out what the acceleration of each of the two blocks is during the motion.

I might be wrong though. I don't have the best physics track record on this forum, lemme tell ya.

But in this problem the static friction between the 2 blocks is important. We have to make sure the top block is not sliding on the lower block. If we consider the 2 blocks as 1 unit, how can we ensure this?

Do the question in parts - first work out the maximum force you can apply to the top block without it sliding. Next, consider the system as a whole, as §outh§tar said.

How can the top block accelerate (net force to the right) while we can still consider STATIC friction? (not kinetic friction)
It accelerates relative to the ground. There is no relative motion between the top block and the lower block, but since the lower block is accelerating, the top one is being dragged along with it.

Ok. The top block has a mass of 4Kg. That means the normal force exerted on it by the bottom block is 39.2N. From this, you can find that the Force of static friction between the blocks is 23.52N. In other words, if you pull any harder, the blocks will slip.

The normal force exerted by the ground is going to be 68.6N. That means the force of kinetic friction is 13.72N. Thus the net force is 68.6 - 13.72 = 54.88N.

Since F=ma, and we know m and F, it is a trivial task to calculate a, and thus time.

2a) Here, the man's hand is touching the rope, which complicates things...Will there be double tension? If so, does it mean the hand touching the rope actually helps him accelerate faster?
When the man pulls on the rope, there is an upward force acting on the man.

2b) Yes, I mean a block on earth weighing 400N.
It would weigh the same when accelerating downward at 2m/s, because force of gravity never changes (provided g is constant)
Yes... I should have asked:
What force does the block apply to the rope when accelerating downward at 2m/s?

How can the top block accelerate (net force to the right) while we can still consider STATIC friction? (not kinetic friction)

Because the bottom block is accelerating to the right at the same rate - static friction applies because the surfaces in question aren't sliding over each other.

Ok. The top block has a mass of 4Kg. That means the normal force exerted on it by the bottom block is 39.2N. From this, you can find that the Force of static friction between the blocks is 23.52N. In other words, if you pull any harder, the blocks will slip.

The normal force exerted by the ground is going to be 68.6N. That means the force of kinetic friction is 13.72N. Thus the net force is 68.6 - 13.72 = 54.88N.

Since F=ma, and we know m and F, it is a trivial task to calculate a, and thus time.

There's a couple of mistakes here, Rubiks. It is actually a very interesting question.

There's a couple of mistakes here, Rubiks. It is actually a very interesting question.
What is my mistake?

You used the wrong force in your last step - you're comparing the normal force and the frictional force, when you want to compare the friction (13.72N) with the maximum pulling force (23.52N), which gives a net force of 9.8N.

Oh, that was dumb of me :o

Right... so what's the acceleration?

The acceleration would be the net force over the mass. So that would be applied force minus kinetic friction: 23.52N - 13.72N. The force is 9.8N. Divide that by the mass (7Kg) to get an acceleration of 1.4m/s^2.

That's the mistake I made too.
The 23.52N is the maximum force that the two blocks can apply on each other, right?

i have provided a solution to this problem is the file that followa..please let me know if there is a flaw in the method or if my handwriting s too cryptic

http://www.uploading.com/files/OM9SP8XY/answer.pdf.html

The acceleration would be the net force over the mass. So that would be applied force minus kinetic friction: 23.52N - 13.72N. The force is 9.8N. Divide that by the mass (7Kg) to get an acceleration of 1.4m/s^2.

that is a wrong approach, since you have considered forces acting only on the 3 kg mass, its acceleration should 9.8/3 and not 9.8/7...because , in the solution you have considered A and B as two seperate systems and not as one single system of mass 7, A remains at rest wrt to B and B accelerates wrt the earth. There for A would have no net force acting on it wrt B.