# Pulley Problem homework

1. Nov 20, 2005

### amcavoy

Here is a link to the pulley diagram:

http://img287.imageshack.us/img287/2904/1711082ao.gif

(a) What is the largest value the force F may have so that m2 remains at rest on the floor?

I said 20N here (which is wrong) because if m2 is on the floor, there will be a downward force on the pulley of 20N. If that was matched by an upward force of 20N it would cancel out.

(b) What is the tension in the string if the upward force F is 100 N?

This would mean that both masses are off the ground. Because of this (relative to the pulley), mass m1 would be accerating upwards at 10 m/s2. I got this by saying -10N + 20N = 10N, then dividing it by one kilogram. However, this is also incorrect.

(c) With the tension determined in part (b), what is the acceleration of m1?

Here I said that the acceleration would be the 10 m/s2 found in part (b) plus 100N/3kg=33.3...m/s2, for a total upwards acceleration of 43.3 m/s2.

All of my answers were incorrect and I cannot see why. Does anyone see my errors? Thank you.

2. Nov 20, 2005

### daniel_i_l

(a) m1 is also pulling m2 up.
(b) why didn't you take the F into account, and tension is a force, not a acceleration.
(c) after finding the tension, fing the total force on m1 and then find the acceleration.

3. Nov 21, 2005

### amcavoy

Hmm... Well I figure that for everything to remain still, the upward force must be 20 N to match the 10N on both sides of the pulley pulling downwards (I'm using g=10 N/kg). Now if more force is applied, the block on the right will stay on the ground and the left block will move upwards. If even more is applied, both blocks will be lifted; the block on the right will be pulling the left block up as the upward force does. I'm just having trouble finding what this minimum value is so that the right block stays at rest on the ground. Any suggestions?

Thanks again.

4. Nov 21, 2005

### mezarashi

This question is apparently really trickey. The conditions you can impose at that minimum force is that the tension T in the wire must be equal to m2g, such that no normal surface force acts on m2. Consequently, due to the pulley effect, this tension is also equal to the tension T seen by m1, which will cause an acceleration.

Now take the force analysis of the entire pulley system, we have 3 external forces: force F, and the 2 gravitational forces acting on m1 and m2 causing a net acceleration of m1.

I believe these are enough equations and conditions to solve.

5. Nov 21, 2005

### amcavoy

When there is no normal force on m2, the mass held by the pulley is essentially 3 kg. Multiplying this by g, I get that the upward force is 30 N. Now this is what I did when the force is 100 N upward:

$$a=\frac{g\left(m_2-m_1\right)}{\left(m_1+m_2\right)}+\frac{100\text{N}}{3\text{kg}}=\frac{110}{3}\frac{\text{m}}{\text{s}^2}$$

The above would actually answer part (c) of the problem, and multiplying that by the mass that has this upward acceleration (m1) would give me the tension in the rope.

I cannot see where I am going wrong. I have tried to take your suggestions into account, but somehow my answers are still coming out incorrect. Does anyone know why this is? Thank you very much .

6. Nov 21, 2005

### siddharth

For part a,
If the tension in the rope is $$T$$, then applying Newton's second law for the pulley gives
$$F=2T$$ (Can you see how I get this equation?)
Also, at the limiting condition, N =0. From this, can you find the value of $$F$$?

7. Nov 22, 2005

### amcavoy

In that case I would say T=(m1)g=10N --> F=20N, but that is incorrect.

8. Nov 22, 2005

### siddharth

Shouldn't it be (m2)g. The mass m2 is on the ground isn't it?

9. Nov 22, 2005

### amcavoy

OK. The tension in the rope must be 20 N, to make the net force on m2 zero. The first thing I think is that if the system is to remain in equilibrium (nothing moves at all), the upwards force must be 20 N (is this correct?) to match the 10 N tension on both sides of the pulley. Now, if I increase this enough, the tension will be 20 N on both sides, which means the upwards force is 40 N. <-- could someone tell me if this is a correct answer for part (a)?

10. Nov 22, 2005

### amcavoy

EDIT: I arrived at answers to this problem. I don't know what I was thinking before, but thanks much for all the help .