Newton's Laws - finding acceleration of block on sled

[jgray]

Homework Statement

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?

Homework Equations

F=ma
Ff = ukFn
Ff = usFn
mb = block's mass
mt = toboggan's mass
Fa = applied force (30N)

The Attempt at a Solution

a. For the block:
Fg = mb = (2)(9.8) = 19.6N
No acceleration in y direction therefore FN = 19.6N

Ff = usFn = (0.58)(19.6N) = 11.37N -- from this I know that the applied force overcomes the max static friction force and the block will indeed move and we need to use kinetic friction. (** Am I correct in including this?)

Ff = ukFn = (0.48)(19.6N) = 9.41N

F=ma = Fa - Ff= ma
30N- 9.41N = 2a
a= 10.30 m/s^2, to the left

b. F= mta =
Ff = mta
9.41N = 4a
a=2.35 m/s^2, also to the left (** Am i correct to use the frictional force of the block here?)

c. F(toboggan) = mta
= 9.41N/4kg
=2.35 m/s^2
Since it is the applied force that is decreasing, I have to get the acceleration of the block to 2.35m/s^2.
F(block)=mba
Fa-Ff=mba
F-9.41=(2kg)(2.35m/s^2)
F= 14.11N

Just looking to get some advice as to if I'm going through this correctly! It confused me when giving me both the static and kinetic coefficient of frictions, and looking online it seems like everyone was doing this question differently. Thanks!

 

[Doc Al]

Ff = usFn = (0.58)(19.6N) = 11.37N -- from this I know that the applied force overcomes the max static friction force and the block will indeed move and we need to use kinetic friction. (** Am I correct in including this?)

You'll need to rethink your reasoning. Hint: If the two blocks are were to move together, what kind of friction would act between their surfaces? What acceleration would they have? What force would the block have to exert on the sled? Is static friction enough? (Your reasoning would make sense if the sled were fixed to the ground. But it's not.)

Ff = ukFn = (0.48)(19.6N) = 9.41N

F=ma = Fa - Ff= ma
30N- 9.41N = 2a
a= 10.30 m/s^2, to the left

OK.

b. F= mta =
Ff = mta
9.41N = 4a
a=2.35 m/s^2, also to the left (** Am i correct to use the frictional force of the block here?)

OK.

c. F(toboggan) = mta
= 9.41N/4kg
=2.35 m/s^2
Since it is the applied force that is decreasing, I have to get the acceleration of the block to 2.35m/s^2.
F(block)=mba
Fa-Ff=mba
F-9.41=(2kg)(2.35m/s^2)
F= 14.11N

What's the maximum force that the block can exert on the sled so that they still move together? Compare with the reasoning outlined above.

 

[Adithyan]

You have done first two questions correctly- but not the third.

For the third question, you can't fix the acceleration as 2.35m/s^{2}[\itex]. Since the force changes, acceleration will also change.And friction also changes from kinetic to static. So you have to assume an acceleration &#039;a&#039; which will be common to both toboggan and the block. <br /> <br /> Now, there are two variables &#039;F&#039; (force at which the accelerations become equal) and &#039;a&#039;. Make two equations and solve them. <br /> <br /> Best of luck!<br /> Write to me if you have any question-Happy to help you!<br /> Regards<br /> ADI

 

[haruspex]

c.If the applied force is gradually reduced, at what value do the block and the toboggan have the same acceleration?

This question strikes me as ambiguous. Are we to imagine multiple trials, from rest, with a slightly smaller force in each trial, or is there a single trial in which the force is gradually reduced while the system is moving? I would guess it's the first, and other responders seem to agree, but it is not clear.

 

[SteveS]

I'm currently working on this question myself.

part a). for the block I used the calculation ∑F = 30N - Ff = 30N - (0.48)(19.6) = 20.6 N, then a = 20.6 N / 2.0 kg = 10.3 m/s^2 to the left

part b). for the acceleration of the sled i used the calculation Ff = ma, since there is no other horizontal force on the toboggan.
Ff = ma , a = (19.6 N)(.48) / 4.0 kg = 2.4 m/s^2 to the left

these values seemed to agree with that had been done before.

Part c. is what confuses me. It's not asking for an acceleration, to me the questions is asking for a force that would allow the block and toboggan to move together. Wouldn't this just be when Ffmax = μsFn = (0.58)(19.6N) = 11.4 N? At the Ffmax, the block and toboggan will be on the verge of movement. If we applied a force of 11.400000000000001 N we would have exceeded the static force of friction and then the block and toboggan would switch to a kinetic force of friction and move nearly together? So i would say a force of 11.4 N would be the force that would allow the block and toboggan to move together at the same acceleration.

 

[Doc Al]

Part c. is what confuses me. It's not asking for an acceleration, to me the questions is asking for a force that would allow the block and toboggan to move together.

Right. But calculating the acceleration might be a useful step towards the answer.

Wouldn't this just be when Ffmax = μsFn = (0.58)(19.6N) = 11.4 N?

Yes.

At the Ffmax, the block and toboggan will be on the verge of movement. If we applied a force of 11.400000000000001 N we would have exceeded the static force of friction and then the block and toboggan would switch to a kinetic force of friction and move nearly together? So i would say a force of 11.4 N would be the force that would allow the block and toboggan to move together at the same acceleration.

Careful! You need to figure out what applied force would allow the static friction to max out, but it's not simply equal to the friction force. (Think about it: If the applied force equaled the max static friction, then wouldn't it the block be in equilibrium? Does that make sense?)

Hint: What's the acceleration?

 

[SteveS]

Ok, taking a look at the acceleration then, since as previously mentioned, to keep the box from sliding the Ffmax = μsFn = ma

The Fn exerted by the toboggan on the box is Fn = mg.

substituting into the previous equation, we get

μs mg = ma, the masses will cancel and we get a = μsg = (0.58)(9.80 m/s^2) = 5.7 m/s^2

using this as the acceleration of the combined mass of the toboggan and box would be F = ma = (6.0 kg)(5.7m/s^2) = 34 N

This value of 34 N seems rather illogical based on the initial 30 N that was given in the question producing such different values for a in parts a and b.

The question states that applied force would be reduced so you would already have passed from static friction to kinetic friction, as you decelerate, would it be possible for the force applied to be less than the static force because the μk is lower than the μs?

I'm thinking that I could return my value for acceleration back to the equation for the box used in part a, replacing the 30 N with the variable Fp.

ΣF = Fp - (.48)(19.6 N) = ma

That would make it Fp = (2.0kg)(5.7 m/s^2) + (.48)(19.6N) = 21 N. This makes much more sense as it would be less than the 30N originally applied to the system.

 

[haruspex]

μs mg = ma, the masses will cancel and we get a = μsg = (0.58)(9.80 m/s^2) = 5.7 m/s^2

Is this in respect of the box or the toboggan? If the block, you've omitted the applied force; if the toboggan, the two m's are not the same.

 

[Doc Al]

Ok, taking a look at the acceleration then, since as previously mentioned, to keep the box from sliding the Ffmax = μsFn = ma

The Fn exerted by the toboggan on the box is Fn = mg.

substituting into the previous equation, we get

μs mg = ma, the masses will cancel and we get a = μsg = (0.58)(9.80 m/s^2) = 5.7 m/s^2

As haruspex has already noted, you are mixing up the masses.

Hint: Deal with the toboggan first.

 

[SteveS]

Ah yes, the Fn uses the mass of the box not the sled so they won't cancel out.

Ok so it should have been:

a = (0.58)(19.6N) / (4.0 kg) = 2.8 m/s^2 for the acceleration of the toboggan.

then going back to the box using the same acceleration;

Fp = (2.0 kg)(2.8 m/s^2) + (0.48)(19.6N) = 15 N

Have I finally got it? I was seriously laying in bed last night for about an hour turning this problem around in my mind.

 

[haruspex]

Fp = (2.0 kg)(2.8 m/s^2) + (0.48)(19.6N) = 15 N

Why are you using the kinetic coefficient here? The frictional forces as experienced by the two bodies must be equal and opposite.

 

[SteveS]

I used the kinetic coefficient here, because I was thinking about what my text said about once over coming the static force you then us the kinetic coefficient. However, looking at it makes sense when i look at my previous work. I need x amount of energy to reach Ffmax and then the energy to accelerate the 2.0kg box at the same acceleration as the toboggan.

so it should then be

Fp = (2.0 kg)(2.8 m/s^2) + (0.58)(19.6N) = 17 N

 

[haruspex]

I used the kinetic coefficient here, because I was thinking about what my text said about once over coming the static force you then us the kinetic coefficient. However, looking at it makes sense when i look at my previous work. I need x amount of energy to reach Ffmax and then the energy to accelerate the 2.0kg box at the same acceleration as the toboggan.

so it should then be

Fp = (2.0 kg)(2.8 m/s^2) + (0.58)(19.6N) = 17 N

Yes.

You can get there a bit faster by first looking at the whole system: a = F/(4.0+2.0).
Now considering the top mass only, a = (F-(2.0)(0.58)g)/2.0