# Kinematics Help!

1. Jun 10, 2006

### vinny380

This question, from a previous A.P physics exam, was given as a homework problem. I am wondering if anybody could tell me if I am on the right track or totally wrong.

Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by mass less strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following.
a. The acceleration of the 4 kilogram block
b. The tension in the string supporting the 4 kilogram block
c. The tension in the string connected to the l kilogram block

(Sorry but I cant seem to get the pic on here- basically on one side of the pulley is a 4kg mass, and on the other side is a 2kg mass that is connected by a string to a 1kg mass)

A. F=M*A
F= (3kg)(9.8)
F=29N
29N=(4kg)(a)
A=7.25m/s^2

B. 29N ????? I basically got this calculation from above when finding the acceleration, but i am not sure if this is what they are looking for

C. Ft=M*A
Ft= (1kg)(7.25)
Ft= 7.25N

For some reason, I am really confused with this problem. Please help!!

2. Jun 10, 2006

### Hootenanny

Staff Emeritus
You may want to draw free body diagrams of all the forces acting on each block. I would make the 2kg and 1kg blocks into a single block of mass 3kg. Don't forget the tension in the strings.

3. Jun 10, 2006

### vinny380

Thats what I did in part A ...combined the 2kg mass and the 1 kg mass to form th 3kg mass....was that not correct?

4. Jun 10, 2006

### Hootenanny

Staff Emeritus
It was indeed correct, however, gravity is not the only force acting on the blocks; you are forgeting the tension in the string.

5. Jun 10, 2006

### vinny380

hmmmmm...then how could i solve for the tension force?? And would that change the answer for A?

6. Jun 10, 2006

### Hootenanny

Staff Emeritus
Yes it would change your answer to part a. You need to sum forces and use Newton's second law;

$$\sum \vec{F} = m\vec{a}$$

Note that the tension in the string in this case is uniform and that the acceleration is also uniform and equal for both particles.

7. Jun 10, 2006

### vinny380

I understand that I have to use F=MA to find all the forces to find the acceleration. But I thought I accounted for that by multiplying 9.8 * 3kg=29N .....would I have to multiply 4kg*9.8.. and then subtract?
So, 39N-29N=10N
10N= (4kg)(a)
10N/4=A
A=2.5ms^2????????

8. Jun 10, 2006

### Hootenanny

Staff Emeritus
Here I'll set up the two equations for you;

(1)$4g - T = 4a$ - The tension is acting in the opposite direction of motion.

(2)$T - 3g = -3a$ - The tension is acting in the same direction of motion.

I really would recommend drawing a free body diagram.

Edit: nrqed seems to explain it better than me

Last edited: Jun 10, 2006
9. Jun 10, 2006

### nrqed

Your F is only the weight of the combined 3 kg. It is NOT the net force on that mass since you are not including the tension. So you cannot use the 29 N to get an acceleration (an acceleration requires the *net* force).

You will have to set up two equations with two unknowns; the unknowns will be the acceleration and the tension. The two unknowns will appear in both equations.

As Hootenanny said, you really have to draw a free body diagram first.

10. Jun 10, 2006

### nrqed

Unfortunately, this is incorrect. The two acceleration terms should have opposite signs (otherwise it would imply that both masses accelerate in the same direction which is impossible) and the tensions must have the same sign (the tension is opposite to the gravitational force for both masses).

I think it is easier to start this way:

For the 3kg mass (call this mass A), we have
$$-(3 kg) g +T = (3kg) (a_y)_A$$

for the second one we have

$$-(4 kg) g + T = (4 kg) (a_y)_B$$

Now, if one mass is accelerated upward, the second mass will necessarily be accelerated downward, so $(a_y)_A= - (a_y)_B$ . then we have

$$-(3 kg) g +T = (3kg) (a_y)_A$$
$$-(4 kg) g + T = - (4 kg) (a_y)_A$$

Last edited: Jun 10, 2006
11. Jun 10, 2006

### Hootenanny

Staff Emeritus
I apologise, but I realised I had made a mistake and edited my post, but it must not have showed up immediatly. Congratulations on your medal btw, well deserved.

Last edited: Jun 10, 2006
12. Jun 10, 2006

### nrqed

Thank you Hootenanny, and thanks for recommending me
(btw, I think you forgot to change the sign of the tension in one equation...the tension is opposite to the weight for both the 3 kg and the 4kg).

Thanks again!

Patrick

13. Jun 10, 2006

### Hootenanny

Staff Emeritus
Again, it is at the small obstacles I fall :grumpy:

14. Jun 10, 2006

### arunbg

Can't wait to see your avatar nrqed . :D

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