# Calculating the frictional forces on a sliding box on a ramp

Karagoz

## Homework Statement

We kick a box upwards a ramp. The box slides upwards, then slides backwards again. We ignore the air drag and other factors. We only look at the gravity, normal forces and friction forces.

Forces on the box when it's moving upwards: Weight of the box is 0.123 KGs.

Angle α is 24°.

Gravitational acceleration is -9.81 m/s^2

Acceleration of the box when it's moving upwards is: -6.9 m/s^2.

Acceleration of the box when it's moving downwards is -1.16 m/s^2

The question is to find the R (frictional force) when the box is moving upwards and downwards.

G_x = mg*sin(α)

## The Attempt at a Solution

I have attemted to solve the problem, and I got almost equal values for R (ca. 0.35N). But in the "answer key" of the problem this is how the problem is solved:

Friction force when box is moving upwards:
F_x = ma
-G_x - R = ma
R = -ma -G_x
R = -ma -mg*sinα
R = -m(a + g*sinα)
R = -0.123kg(-6.9m/s^2 + 9.81m/s^2 * sin24)
R = -0.41N

Friction force when force is moving downwards:
F_x = ma
-G_x + R = ma
R = ma -G_x
R = ma +mg*sinα
R = -m(a + g*sinα)
R = 0.123kg(-1.16m/s^2 + 9.81m/s^2 * sin24)
R = 0.35N

Is the blueprint wrong or?

## Answers and Replies

Homework Helper
R = -0.123kg(-6.9m/s^2 + 9.81m/s^2 * sin24)
R = -0.41N
Plug these numbers into a calculator and see if you get the same result.

Now when the block is moving uphill, all vectors point downhill, so the proper equation to write is (and I put numbers although I find this distasteful, but I have to make the point clear) $$0.123(kg)*[-9.81(m/s^2) \sin24^o ]+(-R_{up})=0.123(kg)*[-6.9(m/s^2)]$$ For the downhill motion, the force of friction changes sign and $$0.123(kg)*[-9.81(m/s^2) \sin24^o ]+R_{down}=0.123(kg)*[-1.16(m/s^2)]$$