In the figure below the two blocks are connected by a string of negligible mass passing over a frictionless pulley. m1 = 10.0 kg and m2 = 3.80 kg and the angle of the incline is = 37.0°. Assume that the incline is smooth. (Assume the +x direction is down the incline of the plane.) For what value of m1 (in kg) will the system be in equilibrium?

For equilibrium, the mass m1 must be about 2.29 kg

Explanation:

The forces acting on m1 are : the weight (m1 * g) and the tension (T1) of the string.

The forces acting on m2 are:

1) along the ramp: the component of m2's weight (m2 g * sin(37)) and the tension (T2) of the string. Which by the way, must be equal in magnitude to T1 since the string is inextensible.

2) perpendicular to the ramp; the component of m2's weight (m2 g * cos(37)) and the normal from the contact with the ramp. These two compensate each other.

We therefore want the net force on each block to be zero for the system to be in equilibrium. This means:

T1 = m1 g

T2 = T1 = m2 g sin(37)

Then we have that if m2 = 3.8 kg, then:

m1 g = m2 g sin(37)

cancelling "g" on both sides:

m1 = m2 * sin(37) = 3.8 * sin(37) = 2.28689 kg

which may be rounded to about 2.29 kg.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.