# Magnitude & Direction of the acceleration of these blocks [Pic of Progress Included]

1. Jul 24, 2012

### ksm2288

1. The problem statement, all variables and given/known data
Block of mass mA = 2kg - lying on frictionless inclined plane with a slope of 20 degrees.

connected to mA via frictionless, massless pulley and massless cord... mB = 3kg

http://img152.imageshack.us/img152/8286/imag0103l.jpg [Broken]

2. Relevant equations

F= Mass * Acceleration

Acceleration = Mass / Force

3. The attempt at a solution

so far all I can get is the free body diagrams..
would this follow a = Fga + Fgb / ma + mb?

except I don't have force..

Last edited by a moderator: May 6, 2017
2. Jul 24, 2012

### azizlwl

Re: Magnitude & Direction of the acceleration of these blocks [Pic of Progress Includ

You have to draw 2 diagrams.
1. Forces exerted on mass#1
2. Forces exerted on mass#2

Make the unknown force as T.

3. Jul 24, 2012

### ksm2288

Re: Magnitude & Direction of the acceleration of these blocks [Pic of Progress Includ

I have drawn two diagrams? I attached a picture. So I'm just calculating the tension force?

4. Jul 25, 2012

### ehild

Re: Magnitude & Direction of the acceleration of these blocks [Pic of Progress Includ

The normal force N is perpendicular to the slope. It is not vertical.
What do you call Fgma and Fgmb?
You drew only one force exerted to mb. You need to draw the force of gravity, too.
Remember that the tension, the magnitude of the force the cord exerts, is the same on both bodies.

The equation is incorrect. You miss a few parentheses.
Explain, what Fga and Fgb are.

ehild

Last edited by a moderator: May 6, 2017