Find the magnitude of the applied force F

[ryan]

A 115.0 kg box is pushed by a horizontal force F at constant speed up a frictionless ramp which makes an angle of 49.0 deg with the horizontal. Find the magnitude of the applied force F.

I can't seem to get this problem right. In the x-direction there is a push. Which I got mgsin(49).
In the y-direction there is the normal force and mg. mg is negative and I think the normal force is mgcos(49). I'm not positive if this is right but when i add -mg+mgcos(49) and do vector addition on the x and y directions I get the wrong answer. What am i doing wrong?

 

[sridhar_n]

Hi

What about the frictional force? Don't u have the value of & mu?




Sridhar

 

[ryan]

the ramp is considered frictionless

 

[HallsofIvy]

Since there are no friction forces you need to do two things:
determine the component of the force in the direction of the incline:
Since the angle of the incline is given as 49 degrees to the horizontal, constructing the right triangle with legs the along the incline and normal to it with hyponetuse horizontal has angle 49 degrees at the bottom and so near side/hypotenuse = cosine. It is correct that the force along the incline would be F cos(49).

I don't know why you would ADD mg and mg cos(49)- that would be adding the total weight to one component. And the WRONG component, by the way: since the weight vector is vertical, not horizontal the component of weight in the direction of the incline is given by the sine: -mg sin(49)= -115(9.81) sin(49)

Of course, we must have F cos(49)= 115(9.81) sin(49). Solve that for F.