Normal force on an inclined plane

[mbrmbrg]

A crate of mass m = 100 kg is pushed at constant speed up the frictionless ramp (theta= 27.0°) by a horizontal force F.
What is the magnitude of the force exerted by the ramp on the crate?

My x-y coordinate system has x+ pointing up the ramp.

I assume that the force in question is the Normal force, which points directly along my positive y-axis.

No acceleration in the y direction (or x direction either, but that's irrelevant for this part of the problem), so
\Sigma\\F_{y}=N-mg_{y}=N-mg\cos\theta=0
which gives N=mg\cos\theta=(100kg)(9.81m/s^2)(cos27^o)=874N

But that's not the right answer... (and I don't know what the right answer is, either)

Help?

 

[radou]

A crate of mass m = 100 kg is pushed at constant speed up the frictionless ramp (theta= 27.0°) by a horizontal force F.
What is the magnitude of the force exerted by the ramp on the crate?

My x-y coordinate system has x+ pointing up the ramp.

I assume that the force in question is the Normal force, which points directly along my positive y-axis.

No acceleration in the y direction (or x direction either, but that's irrelevant for this part of the problem), so
\Sigma\\F_{y}=N-mg_{y}=N-mg\cos\theta=0
which gives N=mg\cos\theta=(100kg)(9.81m/s^2)(cos27^o)=874N

But that's not the right answer... (and I don't know what the right answer is, either)

Help?

Constant velocity implies that the resulting force exerted on the crate has to equal zero. First write the equation of equilibrium for the assumed x-direction, to obtain the magnitude of the force F. Then write the equilibrium equation for the y-direction, to obtain the magnitude of the normal force N exerted on the crate from the ramp. Your mistake was that you did not include the pushing force F in your equation.

 

[mbrmbrg]

Got it; even after I tilted my coordinate system, I assumed that a horizontal force has no y-component. Bad thing.

Thanks!