# Homework Help: Forces & Friction

1. Apr 26, 2014

### Draggu

1. The problem statement, all variables and given/known data

A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on top of the toboggan. The coefficient of static friction μs between the block and the surface of the toboggan is 0.58, whereas the kinetic friction coefficient is 0.48. The block is pulled by a horizontal force of 30 N as shown.

a] Calculate the block’s acceleration.
b] Calculate the toboggan’s acceleration
c] If the applied force is gradually reduced, at what value do the block and the toboggan have the same acceleration?

2. Relevant equations
mBa = 30 - Ff
mTa = Ff

μs * mg = force of static friction
μk * mg = force of kinetic friction

T=toboggan
B=block

3. The attempt at a solution

So for a) I get:

mBa=30-9.408
a=10.3m/s^2

b) mTa = Ff (I'm still trying to figure out why friction reverses direction for the toboggan, but alas):

a=2.352m/s^2

All frictions above were calculated using μk.

c) This one's kinda tricky:

Have to find F when acceleration is equal. Since they accelerate the same, this means the block doesn't move relative to the toboggan, so we're using μs rather than μk.

m(B+T)a = F - μs*mg
m(B+T)a = μs*mg

Set them equal (masses of 6kg cancel out):

F - (0.58)*(2kg)(9.8m/s^2) = (0.58)*(2kg)(9.8m/s^2)
F-11.37 = 11.37
F = 22.74N

2. Apr 26, 2014

### PhanthomJay

what you refer to as a reversed direction force is actually an equal but opposite Newton 3rd Law force acting on a different object . Part c is not correct. You should draw free body diagrams of each mass as well as a free body diagram of the combined mass system. When you look at the toboggan for example, the friction force is the net force acting on the toboggans mass only.

3. Apr 26, 2014

### Draggu

Hi Jay, thanks for replying. I realized this a bit later after I made the post. Here is my new attempt:

mBa = F - μs*mg
mTa = μs*mg

Get an acceleration of 2.8425m/s^2

Which leads to a force of 17.055 N

Reasoning: For the block to stay still on the toboggan, I am just looking for the maximum force where the block/toboggan move together.

The friction that the toboggan experiences is just the opposite direction of the friction the block experiences.

Last edited: Apr 26, 2014
4. Apr 27, 2014

### PhanthomJay

Yes looks good now. You should check your work to see if the calculated force accelerates the 6 kg block and toboggan system together at the calculated acceleration.

5. Apr 27, 2014

### Draggu

Oh, so because we're using static force, then the block is essentially "stuck" to the toboggan, right? In this case, force would be calculated using both the masses, whereas before we only calculated the masses individually because they weren't really moving as a combined system.

In this case,

a = 2.8425m/s^2 (same)

BUT, force is m(B+T)a = F - μs*mg

(6kg)(2.8425m/s^2) = F - (0.58)(2KG)(9.8)
F = 28.408N

I'm not sure if this is what you meant, but if it's wrong, is there an easy way to develop intuition for knowing which masses to use with regards to static/kinetic friction?

Last edited: Apr 27, 2014
6. Apr 27, 2014

### PhanthomJay

It is wrong. When you look at the block and toboggan together as one unit "stuck" together, the friction force between the block and toboggan is internal to the system , so it does not enter into your calculation for the net force. Only when you break the system apart do the internal forces show up as "external" forces. Note however that when accelerations of block and toboggan are different, do not look at the combined system this way.

7. Apr 27, 2014

### Draggu

Oh ok, so essentially the calculation (as originally) would be:

(2kg)(2.8425m/s^2) = F - (0.58)(2KG)(9.8)
F=17.005N

I wasn't really sure what you meant by "You should check your work to see if the calculated force accelerates the 6 kg block and toboggan system together at the calculated acceleration."

8. Apr 27, 2014

### PhanthomJay

We'll if the combined mass is 6 and a is 2.8425, then solve for the net force per Newtons 2nd law to get
F_net = ??? Which should agree with your prior result, yes?

9. Apr 27, 2014

### Draggu

Ah! That's what you meant! I think I understand it now! Thanks for all your help (and you're welcome to those who google this question in the future)

10. Jun 30, 2014

### bluenoser

I don't get the 10.3 answer for a). This would be true for a box sitting on the ground.. The reference would be the ground. But the box is on the toboggan and so now the reference is the toboggan. Once the toboggan starts to move, I would think its acceleration would be added to the boxes acceleration to give an overall acceleration wrt the ground. I can't see how you can ignore the fact that the toboggan is moving when computing the acceleration of the box. It appears that you are ignoring the frame of reference.

11. Jun 30, 2014

### PhanthomJay

When using Newton's 2nd law , the frame of reference is the inertial stationary frame (the ground), that is, using Newton 2, accelerations are with respect to a stationary observer. If the toboggan was firmly anchored to the ground, the acceleration of the block would still be 10.3 m/s^2 with respect to the ground, same result. The acceleration of the block with respect to the toboggan is ___??

12. Jun 30, 2014

### bluenoser

Thank you, PhantomJay for your reply. However, I seem to have a blind spot on this one problem. Others in this area I seem to have no issues with. Do you mind if I probe a bit further?

I understand about the ground reference (I think). and so..

Let's suppose we have our stationary observer. And lets suppose the toboggan is bolted to the ground. Then the block moves at 10.3 m/s^2. Easy.. But.. now the toboggan is unbolted .. and it moves due to the friction 'pull' on it from the box. So the toboggan moves at, say 2m/s^2.

The 30N force is still being maintained on the box. Intuitively I think that the box is still 'slipping' on the toboggan at the same rate, thus it still has 10.3 m/s^2 acceleration wrt the toboggan surface and so the stationary observer would see the toboggan accelerating at 2 and the toboggan at 10.3 (wrt toboggan) and so would see the box accelerating at 12.3.

I'm wondering if the understanding of this problem has something to do with the 30N force. In electricity all voltages have a reference and I'm wondering if forces like this all have a reference. I'm not sure how one measures a 30N force. Do you need to have the measuring system 'tied' to some frame? Thus my way of looking at it would only make sense if the 30 N is measured wrt the toboggan. Is there such a thing in physics? Maybe what I need to wrap my head around is that this 30N force uses the ground as a reference, not the toboggan. If it used the toboggan, would the same 30N force 'measure' differently?

Hope the above makes sense.

13. Jul 1, 2014

### PhanthomJay

In the given original problem, the box accelerates at 10.3 m/s^2 with respect to the ground. The toboggan accelerates at 2.3 m/s^2 with respect to the ground . The box accelerates at 10.3 - 2.3 = 8 m/s^2 with respect to the toboggan, using the vector equation for relative accelerations.
Real forces, whether contact or action at a distance forces, are with respect to the objects on which they act. The applied force F acts on the box. The toboggan knows nothing about the applied force F. All it knows is that it is accelerating forward at 2.3 m/s^2 wrt the ground, due to the friction force from the box on it. Whether the applied force on the box is 30 N or 3000 N, it still accelerates at 2.3 m/s^2 with respect to the ground.