# Horizontal Force on a Ramp required to Accelerate a Mass

## Homework Statement

A 5.57 kg box sits on a ramp that is inclined at 38.0^\circ above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.28.

What horizontal force is required to move the box up the incline with a constant acceleration of 4.30 m/s^2?

## Homework Equations

Calculation F. F=ma is important. Trig functions. Calculation of F caused by friction.

## The Attempt at a Solution

I calculated the force required to move the object up the ramp using a horizontal force; m*a/cos(38), which is 23.951N.

I calculated the force caused by the μk, which is sin 52*mg, which is 12.044 N

I added these, and found the horizontal component required, which is those added/ cos 28, which gives 45.7N

I am fairly certain I'm doing this wrong though, and I would appreciate any help.

Thanks!

Delphi51
Homework Helper
Welcome to PF, GiantCube.
I get the 12 and the 24 but not the 45.7. It would be nice to see your F = ma statement with the three things added to get the total force. We may have some differences in the signs.

Thanks. :)

The Force is the horizontal portion of the acceleration, so (m*a*cos(38)) must the force to get over the friction, so sin(52)*m*g. Which would be F= sin(52)*m*g + (u*m*a*cos(38))

... I'm fairly certain that's wrong, but not sure why...