Sliding Block and Pulley

1. Homework Statement
A 5.50 kg block (m1) is connected by means of a massless rope to a 3.15 kg block (m2). The pulley is frictionless. Calculate the maximum value for the coefficient of static friction, if the 5.50 kg block is to begin sliding.

The diagram is attatched to the post.

2. Homework Equations
Friction coefficient = (Frictional Force)/(Normal Force)
Acceleration = (m2*9.8)/(m1 + m2)
Of course, Force = mass*acceleration

3. The Attempt at a Solution
Acceleration = (m2*9.8)/(m1 + m2)
Acceleration = 3.5688 m/s^2
3.5688*5.5 = 19.6284 N = frictional force
Normal force = 5.5*9.8 = 53.9 N

19.6284/53.9 = .36416

I know that the answer isn't right, what am I doing wrong?

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Doc Al
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Attach the diagram as an image. (That site requires authorization to view.)

Ok, I did.

Doc Al
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Note that the block is just about to barely begin sliding. So what can you take as the acceleration? (Hint: Note that you are finding the coefficient of static friction.)

I already gave the acceleration, 3.5688 m/s^2, and doesn't the force of static friction balance the force of tension, which is 3.5688*5.5?

Doc Al
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I already gave the acceleration, 3.5688 m/s^2,
I know. My point was that this acceleration is wrong. How did you arrive at this value?
and doesn't the force of static friction balance the force of tension,
Exactly! So what's the acceleration if the forces are balanced?

The weight of the hanging mass exerts a net force on both masses. (Assuming that the surface is frictionless)

(3.15*9.8)/(3.15+5.5) = 3.5688 m/s^2

Also, the acceleration needs to be zero, so then i did 5.5*3.5688 to find the frictional force.

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Doc Al
Mentor
The weight of the hanging mass exerts a net force on both masses. (Assuming that the surface is frictionless)

(3.15*9.8)/(3.15+5.5) = 3.5688 m/s^2
You found the acceleration of the masses assuming no friction. But you know there's friction, so this is not relevant.

Also, the acceleration needs to be zero
Doesn't this contradict what you just did above?

The acceleration is zero. So what's the net force on m1?

there is none because force is equal to mass * acceleration. so then what am i doing wrong?

Doc Al
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The net force is zero. So add up the forces on m1 and set equal to zero. What horizontal forces act on m1?

(This is a static equalibrium problem.)

Ok, I got it right...

I figured that the tension in the rope would be equal to the weight of m2 in equilibrium.
Then, I set it to equal the force of friction.
I divided the force of friction by the mass of m1 to get .573.

Thanks for the help!

Doc Al
Mentor
Good! You're welcome.