# Simple force question

1. Sep 26, 2012

### oreosama

1. The problem statement, all variables and given/known data

a load of bricks of mass m hangs from one end of a rope that passes over a small frictionless pulley. a counterweight, twice as heavy, is suspended from the other end of the rope. given m determine the accel of the masses and the tension in the rope

2. Relevant equations
f=ma

3. The attempt at a solution

Fa=-2mg + T

2m*a=-2mg + T

a= (-2mg+T)/2m

plug in accel in other force

Fb = -mg + T

m((-2mg+T)/2m) = -mg + T

-2mg + T = -2mg + 2T

T = 2T

T=0?

i have no idea if I am doing anything remotely close to the right path, if I'm way off please explain like a baby as I'm confused thanks for help

2. Sep 26, 2012

### azizlwl

Fa=-2mg + T

2m*a=-2mg + T

a= (-2mg+T)/2m

plug in accel in other force

Fb = -mg + T

m((-2mg+T)/2m) = -mg + T

-2mg + T = -2mg + 2T
----------------------------
FFrom your calculation for Fa, T is greater than 2mg, so it is going up.
When one side going up, the other side should be going down.

But for calculation for Fb, T is greater than mg. It means going up, where you have assumed that it should be going down.
It is a single cable.

3. Sep 26, 2012

### marvincwl

m(a-g)+T= 2mg-T
ma= 3mg-2T
a=(3g-2(T/m))

this is the acceleration of the mass?

4. Sep 26, 2012

### ehild

You would understand the problem better with a drawing showing the forces on both objects, brick and counterweight, as in the attachment.

In what direction will the brick and counterweight move? What is the net force acting on the brick? What is the net force acting on the counterweight?
How are the accelerations related?

ehild

#### Attached Files:

• ###### brickpulley.JPG
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5. Sep 26, 2012

### oreosama

so since pulley requires each to go in opposite direction you make one force negative relative to coordinate system:

(using fa from above)

fb = -mg + T
-ma = -mg + T

...

T = 4/3mg?

6. Sep 26, 2012

### ehild

T=4/3 mg is correct.

ehild

7. Sep 26, 2012

### oreosama

I don't want to open another thread

A block of ice of mass m is released from rest at the top of a frictionless ramp of length L. at the bottom of the incline the speed of the ice is v1. given m,L,v1 determine

the time needed to hit the bottom
the angle between the ramp and the horizontal
the accel of the ice

I made my coordinate system slant at the angle of the ramp because I thought this would make it less work

http://i.imgur.com/d18GN.png

(is this okay?)

x0 = 0
x = L
v0 = 0
v = v1
a= A
t = T

v1 = 0 + AT

T = v1/A

v1^2 = 2AL

A = v1^2 / 2L

T = v1 / (v1^2 / 2L)

T = 2L / v1

I got these answers pretty cleanly but I have no idea what I should be doing to get theta since I dont see a way to get the lengths of the potential triangle with hypoteneus L

8. Sep 26, 2012

a=gSinθ