# Friction and Pulleys

1. If flat car is given an acceleration a = 3 m/s^2 starting from rest, compute tension (in N) in the light inextensible string connected to block A of mass 30 kg. Coefficient of friction between block and flat car is = 0.50.

Neglect mass of pulley and its friction. Take g = 10.

2. Friction = ## \mu N ##
F = ma
ma = mg - T

3. No idea. I couldn't approach the problem. Please tell me how to approach these types.

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pcm

1. If flat car is given an acceleration a = 3 m/s^2 starting from rest, compute tension (in N) in the light inextensible string connected to block A of mass 30 kg. Coefficient of friction between block and flat car is = 0.50.

Neglect mass of pulley and its friction. Take g = 10.

2. Friction = ## \mu N ##
F = ma
ma = mg - T

3. No idea. I couldn't approach the problem. Please tell me how to approach these types.
Write the newton's laws in the frame of the ground. It is an inertial frame. Write down all the forces(along both axes). Also observe that for every meter the car moves, block A moves by 2 meter.

Take g = 10.
I hope the original problem gave some units for g!

haruspex
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In case pcm's post is not clear enough, it's the forces on the block that matter. You don't need to analyse forces on the cart since its motion is a given.

haruspex
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What are the horizontal forces acting on the block? What is the acceleration of the block? What equation does that give you?

What are the horizontal forces acting on the block?
Friction and tension.

What is the acceleration of the block?
Same as acceleration of the car(truck)?

What equation does that give you?
Net force on the block=mass of the block multiplied by acceleration of the truck.

if the acceleration of the block = 2 * truck acceleration, then the force required in the string = m * a

acceleration force is forward (say + ve) and friction force is rearward (say - ve), the tension force is the difference.

haruspex
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acceleration force is forward (say + ve) and friction force is rearward (say - ve), the tension force is the difference.
Yes, it produces the same equation, but I feel it would be less confusing for gracy to stick to the normal formulation ##\Sigma F_x = m a_x##.
gracy, do you see why the acceleration of the block is not the same as the acceleration of the truck?

acceleration force is forward (say + ve) and friction force is rearward
But according to
The page number 6 of attached pdf
the direction of friction force should be forward.

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haruspex
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But according to
The page number 5 of attached pdf
the direction of friction force should be forward.
That's a different situation. In the textbook example, the only force accelerating the block is the friction, so of course it must be acting in the forward direction. In this thread, there is a string pulling on the block.
Friction always opposes relative motion of the two surfaces in contact. Which way does the block move relative to the truck?

Which way does the block move relative to the truck?
rearward.

if the acceleration of the block = 2 * truck acceleration
Why?

haruspex
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rearward.
Yes, sorry, I meant in this thread, which way does it move relative to the "flat car".

haruspex
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Why?
What is the acceleration of the block relative to the car?
Does the total length of the string change?

which way does it move relative to the "flat car".
After Considering pull on block by string?

Does the total length of the string change?
No,it is for sure.

haruspex
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After Considering pull on block by string?
Of course.

Tension in string is pulling the block in forward direction, right?

haruspex