What is the acceleration of the crate?

[Sylvia]

Problem: A 22.0-kg crate is pulled along a horizontal floor by the ideal arrangement
shown in figure below. The force F is 300 N. The coefficient of friction between the crate and
the floor is 0.270. What is the acceleration of the crate?

Homework Equations

F = ma
Friction = (coefficient)(normal force)[/B]

The Attempt at a Solution

I tried F - Friction = ma, but I did not get the correct answer. Please help. [/B]

 

[ehild]

Welcome to PF! :)

F acts on the pulley, not on the crate. The crate is connected to a rope. What is the force the rope pulls the crate with?
Note that the pulley is ideal, massless. What forces act on it, and what should be the resultant force?

 

[Sylvia]

Thank you! :)
That's the tension force, right? So would I set it up as F + T - Friction = ma?
I also wondered this at first, but I got confused because how would you know what the value of T is?

 

[haruspex]

That's the tension force, right?

Yes.

So would I set it up as F + T - Friction = ma?

No. Consider the pulley and the crate separately. What forces act directly on the crate? (Hint: F does not.) What forces act on the pulley?
You can write a ##\Sigma F_x = m a_x## equation for each, but in the case of the pulley the mass is zero.

 

[ehild]

Thank you! :)
That's the tension force, right? So would I set it up as F + T - Friction = ma?
I also wondered this at first, but I got confused because how would you know what the value of T is?

F does not act on the crate. F is applied at the pulley. So what equations you have both for the crate and for the pulley?

 

[Sylvia]

Yes.

No. Consider the pulley and the crate separately. What forces act directly on the crate? (Hint: F does not.) What forces act on the pulley?
You can write a ##\Sigma F_x = m a_x## equation for each, but in the case of the pulley the mass is zero.

See, that's where I'm confused. For the crate would I have -Friction + Tension = ma? And then for the pulley I would have F + Tension = ma? I'm not sure where the tension is playing a role.

 

[ehild]

See, that's where I'm confused. For the crate would I have -Friction + Tension = ma?

Correct.

And then for the pulley I would have F + Tension = ma? .

Think: There are two pieces of the string pulling the pulley. The tension is the same in both pieces. And what is the direction of force one string exerts on the pulley?
And the pulley is massless.

 

[Sylvia]

Correct.

Think: There are two pieces of the string pulling the pulley. The tension is the same in both piece.

Since the mass of the pulley is 0 and since the tension in both pieces is the same, I would get F - 2T = 0?

 

[ehild]

Since the mass of the pulley is 0 and since the tension in both pieces is the same, I would get F - 2T = 0?

Exactly! You can solve the problem now. :)

 

[Sylvia]

Thank you very much for the help!

 

[ehild]

You are welcome. What did you get for the acceleration?

 

[Sylvia]

I actually had a different version of this question; I googled the question and found this version so I could copy and paste it (the question was the same, just the numbers were different). I had 24 kg and coeff. of friction was 0.290. But I ended up getting the right answer, which was 3.41 m/s^2 :)

 

[ehild]

Correct! Good work.:)