Solving Mechanics Problem with Pulleys: Acceleration of 2.0 kg Block

[physicsboy007]

b]1. Homework Statement [/b]
There is a 1kg block on top of a 2 kg block. The 2 kg block
A rope pulls on the lower block with a tension force of 20N. The coefficient of kinetic friction between the lower block and the surface is 0.16. The coefficient of kinetic friction between the lower block and the upper block is also 0.16. What is the acceleration of the 2.0 kg block?

Homework Equations

For block 1 (m1):
T-um1g=1a

For block 2 (m2):
F-T-o.16(m1+m2)(g)-0.16(m1)(g)=2a

The Attempt at a Solution

T-0.16(1)(9.8)=a
20-T-0.16(4)(9.8)=2a

a= (13.728-T)/1 a=T-1.568

I eventually get 4.05 but the correct answer is 6.9. Please help, thank you :)

 

[gneill]

First work out what frictional forces to expect. Suppose that the 2kg block was moving and the 1kg block not keeping up, so it's sliding backwards relative to the bottom block (the 2kg block is "slipping" between the upper block and the floor).

What is the magnitude of the friction force developed between the 2kg block and the floor?
What is the magnitude of the friction force developed between the 1kg block and the 2kg block?

 

[physicsboy007]

I've included those in my equation already...

themagnitude of the friction force developed between the 2kg block and the floor?
= o.16(m1+m2)(g)

the magnitude of the friction force developed between the 1kg block and the 2kg block?
=0.16(m1)(g)


2kg block has 1 kg on top so I added both masses.

 

[gneill]

Okay, so what is the total force acting on the 2kg block in the horizontal direction?

 

[physicsboy007]

Is it not
F-T-o.16(m1+m2)(g)-0.16(m1)(g)=2a ?

20-T-0.16(3kg)(9.8) - 0.16(1)(9.8)=2a

 

[gneill]

I would say that:

Letting g = 9.807 m/s^2

F1 = 0.16(1kg + 2kg)9.807m/s^2 = 4.707N

F2 = 0.16(1kg)9.807m/s^2 = 1.569N

Fnet = 20N - F1 - F2 = 13.724N

 

[physicsboy007]

Why is it not:
Fnet= T-13.728N=ma

There is still tension pulling the block

 

[gneill]

Isn't the tension the 20N supplied via the rope?

 

[physicsboy007]

I uploaded a picture above, please look. There is a 20 N force from the right and tension to the left

 

[gneill]

Ah! That's a horse of a different color! So the 1kg block is providing more than just a frictional force, it's also adding its inertial force, and the friction between the blocks counts twice!

The net inertia is 3kg. F1 and F2 are still as described previously. Then the net acceleration will be:

(20N - F1 - 2*F2)/3kg = 4.05 m/s^2 as you found.

 

[physicsboy007]

but that is not the right answer, it is 6.9 according to my answer sheet

 

[gneill]

Consider the possibility that the answer sheet may be wrong.

Certainly the inertial mass that is being moved by the 20N applied force is 3kg -- the taught rope joining them via the pulley makes sure of that. The forces acting counter to the 20N are easily enumerated as the frictional forces developed between the sliding surfaces. What's left?

Let's do a reality check. Suppose that the answer was 6.9 m/s^2. Then in order to accelerate 3kgs of mass at that rate the net force would have to be 3kg*6.9m/s^2 = 20.7N. That's more force than is being supplied.

 

[physicsboy007]

Thank you :)