Did I do this question correctly?

[lanvin]

Question is:
"A box with a mass of 22kg is at rest on a ramp inclined at 45° to the horizontal. The coefficients of friction between the box and the ramp are µ(s) = 0.78 and µ(k) = 0.65

Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest."


What I was thinking, is that if the object is at rest, then the net force on the x-axis is zero. So, the magnitude would be the same as the gravity parallel to the ramp...? My calculations:

(22kg)(9.8N/kg)sin45° - F(applied) = 0N
F(applied) = 152N

Is that right, or wrong? Hints or suggestions...?

 

[naresh]

You did not include the effect of friction.

Also, you should be clear when you say "x axis". What is the x-axis in your solution?

 

[PhanthomJay]

Question is:
"A box with a mass of 22kg is at rest on a ramp inclined at 45° to the horizontal. The coefficients of friction between the box and the ramp are µ(s) = 0.78 and µ(k) = 0.65

Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest."


What I was thinking, is that if the object is at rest, then the net force on the x-axis is zero. So, the magnitude would be the same as the gravity parallel to the ramp...? My calculations:

(22kg)(9.8N/kg)sin45° - F(applied) = 0N
F(applied) = 152N

Is that right, or wrong? Hints or suggestions...?

You forgot that the friction force on the box is also acting down the plane.

 

[lanvin]

152N - F(f)
152N - (0.65)(22kg)(9.8N/kg) = 11.86N

did I do it right?

 

[PhanthomJay]

152N - F(f)
152N - (0.65)(22kg)(9.8N/kg) = 11.86N

did I do it right?

No. Why did you use the kinetic friction coefficient when it is given that the box is to remain at rest? Also, the normal force acts perpendicular to the incline; it is not equal to its weight. Have you defined your x axis, and identified all the components of the forces acting on the box along and perpendicular to the incline?

 

[KOKA]

Without any added parallel or perpendicular forces, the force of gravity acting parallel to the ramp will move the box downwards. In this case, you're looking for a larger normal force to keep the box in place (think about adding pressure from a hand to prevent the box from slipping). Remember that F = uFN? Add an unknown force to the FN to symbolize the applied force added to the top of the box.