the bucket moving after it has fallen 1.50 m (assuming that the box has not yet

Question

the bucket moving after it has fallen 1.50 m (assuming that the box has not yet reached the edge of the roof? 24. Two boxes connected by a light string that passes over a light, frictionless pulley. one rests on a frictionless ramp that rises at 30.0° above the horizontal 30.0 kg 30.0 (see Figure 5.52), and the system is released from rest. (a) Make A FIGURE Problem 24. 5 free-body diagrams of each box. kg box move, up the plane or down the (b) which way will t 50.0 plane? Or will it even move at all? Show why or why not. (c) Fin the acceleration of each box.

Explanation / Answer

Let m1 be the mass of the box that is on the incline
let m2 be the mass of the box that is hang up

b)
which direction both boxes move depends on Fweigt(x) of the m1 and the weight of m2

weight(x) = 50*9.8 sin30
weight(x) = 249

weight = 30 * 9.8
weight = 294

The weight of mass 2 is more than the Weight(x) of mass 1, so the both move to the right. Or the 50-kg box move up the incline

c)


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
30*9.8 - [ 50 * a + (50 * 9.8)sin(30) ] = 30 * a
294 - [ 50a + 245] = 30a
294 - 50a - 245 = 30a
49 = 80a
a = 0.6125 m/s^2

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