A 6.0kg box is on a frictionless 45∘ slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.4kg weight.

Question

What is the tension in the string if the 6.0kg box is held in place, so that it cannot move?

If the box is then released, which way will it move on the slope?

What is the tension in the string once the box begins to move?

Answer 1

1.)

If the box doesn't move, then the Tension in the rope is equal to Fg of the 2.4Kg mass.

Fm1=m*a=Fg-T

a=0m/s^2

m*0m/s^2=(2.4kg*9.8m/s^2)-T

T=Fg=(2.4kg*9.8m/s^2)=23.5N

2.)

Fm1=m1*a=T-Fg

2.4kg*a=T-23.52N

Fm2=m2*a=Fd-T

Fd=mg*Sin45=(6.0kg*9.8m/s^2)*(0.70711)

Fd=41.58N

Fd>Fg for m1, so the box will move down the ramp, not up.

3.)

Fm2=m2*a=Fd-T

Fd=41.58N

Fm2=6.0kg*a=41.58N-T

and

Fm1=m1*a=T-Fg

Fm1=2.4kg*a=T-23.52

Solve for a:

a=(T-23.52)/2.4kg

Plug into equation for m2 and solve for T:

6.0kg*[(T-23.52)/2.4kg]=41.58N-T

(6.0T-141.12)/2.4kg=41.58N-T

2.5T-58.8=41.58N

2.5T=41.58N+58.8N

2.5T=41.58N+58.8N

2.5T=100.38

T=40.2N

Answer 2

ERROR IS PREVIOUS ANSWER RESOLVED HERE
In the third part of Devron's equations they dropped the minus T in going from

(6.0T-141.12)/2.4kg=41.58N-T

to

2.5T-58.8=41.58N

So it should be 3.5T-58.8=41.58N