# Need a little help

1. Jun 11, 2006

### MD2000

Hey guys,

I got a couple physics problems I've been workin on but can't seem to figure out..I was wondering if you guys could help me out..

1) The pulley system in the figure is used to lift a crate of mass m = 45 kg. Note that a chain connects the upper pulley to the ceiling and a second chain connects the lower pulley to the crate. Assume that the masses of the chains, pulleys, and ropes are negligible. Determine the tension in each chain when the crate is being lifted with constant speed.

For this problem I figured the force being pulled would be 45x9.8 or 441N..divided by 3 for the three strings would mean 147N each? This is wrong though..any idea what I'm doing wrong?

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2)At the airport, you pull a 13 kg suitcase across the floor with a strap that is at an angle of 40° above the horizontal. Find the tension in the strap, given that the suitcase moves with constant speed and that the coefficient of kinetic friction between the suitcase and the floor is 0.31.

In this problem I divided the problem into x and y components. Since there is a constant speed..a is gonna be 0 and then I pretty much solved for Fn in one of the equations and plugged that back into the other equation solving for the F that is pulling on the string. For this guy I got 28.03..but once again its wrong..

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3) 7. Find the acceleration of the masses shown in the following figure, given that m1 = 0.8 kg, m2 = 1.6 kg, and m3 = 2.6 kg. (Assume the table is frictionless.)

As for this guy I figured the force pulling on m3 is going to be 2.6x9.8 which is 25.48..I then figured that is the force that would pull on the other masses so I did F=(m1+m2)a and got 10.62 but this wasn't correct either..

Any help with any of these guys?

Last edited: Jun 11, 2006
2. Jun 11, 2006

### Mollet1955

Today is 11th, you prepare to leave ?

3. Jun 11, 2006

### Hootenanny

Staff Emeritus
For question two, don't forget that the vertical component will have an effect on the normal reaction force FN;

$$F_{N} = mg - F\sin40$$

Also, donot forget to take into account the frictional force.

For Question three, you are almost there, but you are forgetting that the tension in the rope is acting against m3 so;

$$m_{3}g - T = m_{3}a$$

$$T = (m_{1}+m_{2})a$$

Do you follow?

Last edited: Jun 11, 2006
4. Jun 11, 2006

### MD2000

Ok I think I got three..is it 5.096?

As for two..what I did was:

X: Fcos40=uk x Fn
y: FN = mg-Fsin40

so..

(Fcos40/uk) - mg + Fsin40 = 0
or
F(cos 40 + sin 40) = mg x uk

Solving for F i still get 28.03

5. Jun 11, 2006

### Hootenanny

Staff Emeritus
That looks good to me
Good
Not so good. You have made a manipulation error. It should look something like this;

$$F\cos40 = \mu(mg - F\sin40)$$

$$F\cos40 = \mu mg - F\mu\sin40$$

$$F\cos40 + F\mu\sin40 = \mu mg$$

$$F = \frac{\mu mg}{\cos40 + \mu\sin40}$$

I would also like to complement you on your presentation.

Last edited: Jun 11, 2006
6. Jun 11, 2006

### MD2000

Ahh..stupid math. I'm gettin better with the physics but the math still gives me trouble..is the answer 40.91?

As for the pulley system..its asking for the tension in the rope..when I divided the force by three..I got the force on each rope..since each chain is holding two ropes does that mean that the force on the chain equals twice the force on the rope?

Btw..thanks for all your help I really appreciate it.

7. Jun 12, 2006

### Hootenanny

Staff Emeritus
For your pulley system you know the sum of the foces is zero. Now, in this case the tension is constant throughout the rope. So, if a force F is applied the net force on the left pulley is 2F (both forces acting upwards) do you follow?

8. Jun 12, 2006

### MD2000

So does that mean 2F - mg = ma? Or since the v = 0..2F = mg?

9. Jun 12, 2006

### Hootenanny

Staff Emeritus
The latter is (almost correct), since it states in the question that the crate is being lifted at constant speed a = 0, thus $\sum\vec{F} = 0$. I am assuming you mean a = 0

10. Jun 12, 2006

### MD2000

hehe..yea..i meant a = 0. So..

2T = 45 x 9.8
T = 220.5

So each chain has Tension of 220.5 N?

11. Jun 12, 2006

### Hootenanny

Staff Emeritus
No, to be honest, this question is something of trick question. What you have calculated there is the tension in the rope. Now you need to sum these forces. Now, the pulley attached to the crate is supporting the weight of the crate, therefore the tension is simply mg. Now, the pulley attached to the ceiling has to oppose the tension in the string. Now if we sum the forces acting on the top pulley we see that; $\sum\vec{F} = 2T$ which leads us to a tension of ....

12. Jun 12, 2006

### MD2000

uh wow..thats simpler then I thought..thanks for all your help Hootenanny..I really appreciate it!

13. Jun 13, 2006

### Hootenanny

Staff Emeritus
My pleasure