Normal force on an inclined plane

In summary, the force exerted by the ramp on the crate is equal to the weight of the crate multiplied by the cosine of the angle between the ramp and the x-axis.
  • #1
mbrmbrg
496
2
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
[itex]\Sigma\\F_{y}=N-mg_{y}=N-mg\cos\theta=0[/itex]
which gives [itex]N=mg\cos\theta=(100kg)(9.81m/s^2)(cos27^o)=874N[/itex]

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

Help?
 
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  • #2
mbrmbrg said:
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
[itex]\Sigma\\F_{y}=N-mg_{y}=N-mg\cos\theta=0[/itex]
which gives [itex]N=mg\cos\theta=(100kg)(9.81m/s^2)(cos27^o)=874N[/itex]

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.
 
  • #3
Got it; even after I tilted my coordinate system, I assumed that a horizontal force has no y-component. Bad thing.

Thanks!
 

FAQ: Normal force on an inclined plane

1. What is normal force on an inclined plane?

The normal force on an inclined plane is the perpendicular force exerted by the plane on an object placed on it. It is equal in magnitude and opposite in direction to the component of the weight of the object that is perpendicular to the plane.

2. How is normal force calculated on an inclined plane?

The normal force on an inclined plane can be calculated using the formula FN = mgcos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of inclination of the plane.

3. Why does normal force change on an inclined plane?

The normal force changes on an inclined plane because the angle of inclination affects the component of the object's weight that is perpendicular to the plane. As the angle increases, the normal force decreases and vice versa.

4. Is normal force always equal to the weight of an object on an inclined plane?

No, the normal force is not always equal to the weight of an object on an inclined plane. The normal force only becomes equal to the weight of the object when the angle of inclination is 0°, i.e. when the plane is horizontal.

5. How does normal force affect the motion of an object on an inclined plane?

The normal force on an inclined plane affects the motion of an object by balancing out the component of the object's weight that is perpendicular to the plane. This allows the object to remain in equilibrium or move with a constant velocity along the plane.

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