# Tension/pulley problem, Newtons second law

## Homework Statement

Two objects A and B, with masses m1=1.00kg and m2=2.00kg, are connected with an ideal string. Object A is moving on an ideal incline with 0 friction. Object B is moving vertically. A force F is working on object B. Its magnitude is 6.00N. Object b is accelerating downward, with a magnitude og 5.50 m/s^2.

a) Draw a free body diagram for the two objects
b) Find the tension force between the objects
c) Find the angle beta

## Homework Equations

Newtons second law

## The Attempt at a Solution

Made a free body diagram for each object. Then applied Newtons second law in the x-direction of object A and in the y-direction og object B.
A:
sigma F_x=F-T-m1*sin(beta)=m1*a
sigma F_y=n-m1*g*cos(beta)
B:
sigma F_y=f+T-m2*g=ma

Solved this for the variable T, and ended up with t=20.6 N. Obviously I am doing something wrong here, because when I try to find the angle beta, I end up with cos(beta)>1.

I am not sure whether the force F should be in the free body diagram og object A at all.

Forgot to upload this earlier. :)

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

Two objects A and B, with masses m1=1.00kg and m2=2.00kg, are connected with an ideal string. Object A is moving on an ideal incline with 0 friction. Object B is moving vertically. A force F is working on object B. Its magnitude is 6.00N. Object b is accelerating downward, with a magnitude og 5.50 m/s^2.

a) Draw a free body diagram for the two objects
b) Find the tension force between the objects
c) Find the angle beta

## Homework Equations

Newtons second law

## The Attempt at a Solution

Made a free body diagram for each object. Then applied Newtons second law in the x-direction of object A and in the y-direction og object B.
A:
sigma F_x=F-T-m1*(sin(beta)=m1*a
Does F show up in this free body diagram? And since the block is accelerating down the plane, you have your signage wrong. And you forgot to include the 'g' for the weight component down the plane.
sigma F_y=n-m1*g*cos(beta)
= ? (no acceleration perpendicular to the plane, but this equation is not needed in this problem)
B:
sigma F_y=f+T-m2*g=ma
signage error....net force is down....choose down direction as positive.
I am not sure whether the force F should be in the free body diagram og object A at all.