Question

Two boxes (m1 = 70.0kg and m2 = 51.0kg ) are connected by a light string that passes over a light, frictionless pulley. One box rests on a frictionless ramp that rises at 30.0? above the horizontal (see the figure below), and the system is released from rest. Find the acceleration of each box.

Answer 1

There are 3 type of forces that act on the box that is on the incline (m1). Tension, Weight(x) and Weight(y)

Tension pulls the box up the incline
Weight(x) pulls the box down the incline
Weight(y) pulls the box perpendicular to the incline

Since the there isn no friction on the incline, we do not have to worry about Weight(y)

For mass 1:

Weight(x) = (m1 *g )sin(30)

Fnet = ma

Since the Tension is upward at 30 degrees and Weight(x) is downward at 30 degrees, let Weight(x) be negative

Tension - (m1 * g) sin(30) = m1 * a

Now lets take a look at m2. There are two types of force that act on it. Tension and Weight

Since we already call Tension postive for m1, we make negative Tension for m2 because if they moves, they move in one direction.

For mass 2:

Weight = m2 * g

Fnet = ma
m2 * g - Tension = m2 * a

Now you have 2 equations
m2 * g - Tension = m2 * a
Tension - (m1 * g)sin(3) = m1 * a

Occording to newton's 3rd law. For every action, there is an opposite and equal reaction. The Tension force that acts on m1 is equal the Tension force that acts on m2

Sove for Tension
Tension - (m1 * g)sin(3) = m1 * a
Tension = m1 * a + (m1 * g) sin(30)

plug it it for Tension in the equation of m2
m2 * g - Tension = m2 * a
m2 * g - [ m1 * a + (m1 * g)sin(30) ] = m2 * a

plug in the numbers
51*9.8 - [ 70 * a + (70 * 9.8)sin(30) ] = 51 * a
499.8 - [ 70a + 343] = 51a
499.8 - 70a - 343 = 51a
156.8 = 121a
a =1.29 m/s^2