# What is the applied force required to push an object up a ramp?

## Homework Statement

A cabinet is pushed up a ramp at angle of 10 degrees. The cabinet weighs 100 kg. The friction coefficient is 0.30. What is the minimum applied force required to push the cabinet up the ramp? Calculate the time it takes to push the object up the ramp. The ramp is 15 meters long and elevated to 2.6 meters.

## Homework Equations

All I know how to do is calculate Fdown which is Fdown = Fg(sin10) which is 170 N. Would I add 170 to 0.30 to find the applied force?

## The Attempt at a Solution

See above.

Your help is much appreciated. :)

tiny-tim
Homework Helper
Welcome to PF!

Hi Ken! Welcome to PF! The kinetic friction force is the normal force times µ.

(and remember that the applied force is horizontal, so that will affect the normal force )

Remember that the coefficient of friction is a pure number ratio between the normal reaction to the weight of the object on a ramp and the frictional force needed to stop it moving =]

in other words:

$$\mu \ge \frac{F_{f}}{F_{n}}$$

I figured the applied force needs to be 461 N.

Now I just have absolutely no clue how to calculate the acceleration of the cabinet up the ramp and how to calculate the time. What equation can I use for that?

Thanks again guys. I really appreciate it.

tiny-tim
Homework Helper
A cabinet is pushed up a ramp at angle of 10 degrees. The cabinet weighs 100 kg. The friction coefficient is 0.30. What is the minimum applied force required to push the cabinet up the ramp? Calculate the time it takes to push the object up the ramp. The ramp is 15 meters long and elevated to 2.6 meters.
Now I just have absolutely no clue how to calculate the acceleration of the cabinet up the ramp and how to calculate the time. What equation can I use for that?

Hi Ken! I don't understand that part of the question if "minimum applied force" is used, the acceleration is zero, and the speed can be infinitesimally small, and so the question has no answer. If two coefficients of friction were given, µs and µk, I would assume that the question meant that the force was the "minimum applied force" to get the cabinet started, ie to overcome µs, and then of course once it has started moving, it only has to overcome µk, and so does accelerate.

But only one µ is given. Thanks tiny-tim. :)

Is there a way to find the time it takes to push the cabinet up the 15m if a minimum applied force is used of 461 N?

Hi Ken! I don't understand that part of the question if "minimum applied force" is used, the acceleration is zero, and the speed can be infinitesimally small, and so the question has no answer. If two coefficients of friction were given, µs and µk, I would assume that the question meant that the force was the "minimum applied force" to get the cabinet started, ie to overcome µs, and then of course once it has started moving, it only has to overcome µk, and so does accelerate.

But only one µ is given. I think the coefficient of friction in this case is just for whilst the cabinet is in motion on the ramp, as the question does not state initial rest.

Through drawing a free-body force diagram I have come up with very different answers from the OP.(although likely wrong)

The normal reaction to the curve can easily be given in non-vector form as $$F_{n} = mgcos \theta$$ and by subbing in values: $$F_{n} = 100 \times 9.81 \times cos(10) = 966.1N$$

Then, by finding the frictional force through $$\mu \ge \frac{F_{f}}{F_{n}}$$ and rearranging we get $$F_{f_{max}} = 0.3 \times 966.1 = 289.83N$$

Through resolving the components of the force(289.83N) we now know is pushing up the ramp,$$F_{m}$$ (as there is movement, friction must be at max) we can get the horizontal component of this, $$F_{x}$$ to be $$F_{x} = F_{m}} - F_{n}sin10 - F_{f_{max}}sin80$$

Hi Cilabitaon,

So I calculated Fx to be -167. Now how do I use that number to calculate the time it takes? Sry I'm a physics noob. :) Thanks.

Well, upon further review I have just spouted a load of crap in that last post, sorry =[

Well would Fnet not equal Fpush - Fdown - fk? If I knew what the Fnet was I could determine the acceleration.

If the applied force is 500N(i've changed it from 461N) and Fdown is 171N and fk is 290N then would Fnet not equal 39 N?

Thanks everyone. :)