Box being pushed up an inclined plane

In summary, to determine the magnitude of the normal force and the coefficient of kinetic friction in a scenario where a 250N force is used to push a 29kg box up a 27-degree inclined plane at a constant speed, you must first resolve the forces into the same direction. The normal force will be equal and opposite to the component of the weight of the box acting perpendicularly through the plane. The components of the weight can be found by drawing an angle formed by the vector of weight and a line perpendicular to and under the surface. The normal force will be at 90 degrees to the inclined surface and the x-axis should be parallel to the surface. The options given for the normal force and coefficient of kinetic friction are:
  • #1
bajangal1
5
0

Homework Statement



A 250N force is directed horizontally to push a 29kg box up a plane that is inclined at 27 degrees to the horizontal. the box moves up the plane at a constant speed. Determine the magnitude of the normal force, and the coefficient of kinetic friction.
g=9.8m/s^2
a(y) = 0m/s

Homework Equations


F=mg to find the force that is pulling down on it
Fk= [itex]\mu[/itex] k Fn


The Attempt at a Solution



I understand that the forces need to be resolved into the same direction, but I'm not sure what direction the normal force would be pointing in. The force of gravity i got was 284.2N. I can't figure out the kinetic friction until i figure out the normal force.
 
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  • #2
The normal force will be at 90 degrees to the inclined surface, and equal and opposite to the component of the weight of the box acting perpendicularly through the plane.
 
  • #3
maybe I'm resolving my vectors wrong. when i resolve the vector for weight, the part that should be equal and opposite to Fn I'm getting 129.02N which is not possible as it is multiple choice and that's not even one of the answers
 
  • #4
What are the options you are given?
 
  • #5
a) 330N 0.31
b) 310N 0.33
c) 250N 0.27
d) 290N 0.30
e) 370N 0.26

The answer is e
 
  • #6
To find the components of the weight, draw an angle formed by the vector of weight and a line perpendicular to and under the surface. This angle is equal to the angle of inclination of the plane.
Make the x-axis parallel to the surface, so the y-axis would be perpendicular to the surface and parallel to the normal force.
 

1. How does the angle of the incline affect the force needed to push the box up the inclined plane?

The steeper the incline, the more force is needed to push the box up the plane. This is because the weight of the box is acting more perpendicular to the surface, making it harder to overcome with the force of friction.

2. Is it easier to push a lighter box or a heavier box up an inclined plane?

A lighter box will be easier to push up an inclined plane because it has less weight to overcome with the force of friction. However, the angle of the incline and the coefficient of friction also play a role in determining the ease of pushing the box.

3. What is the relationship between the coefficient of friction and the force needed to push the box up the inclined plane?

The coefficient of friction is a measure of how rough the surfaces are in contact. A higher coefficient of friction means there is more resistance to motion, so a greater force is needed to overcome it and push the box up the inclined plane.

4. Why is it easier to push the box up the inclined plane than to lift it straight up?

When pushing the box up an inclined plane, you are using your force to counteract the weight of the box and the force of friction. Lifting the box straight up requires you to overcome the full weight of the box, making it more difficult.

5. How does the length of the inclined plane affect the force needed to push the box up?

The longer the inclined plane, the shallower the angle will be, and therefore less force will be needed to push the box up. This is because the weight of the box is spread out over a larger distance, reducing the amount of weight acting perpendicular to the surface.

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