- #1
inner08
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A box with a mass of 22kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficient of friction between the box and the ramp are us = 0.78 and uk = 0.65.
a) Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.
b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.
For a, this is what I did:
Fapp = Fs + Fg sin theta
= Usmgcos theta + mg sin theta
= mg (Uscos theta + sin theta)
= (22kg) (9.8m/s^2)(0.78 cos 45 degrees + sin 45 degrees)
= 270N
The largest force that can be applied upward if the box is to stay at rest is 270N.
For b..i'm not totally sure but this is what I got:
Fs = Fg sin theta
= mg sin 45 degrees
= (22) (9.8) (0.707)
= 152N
Fn = Fs * Us
= 152 * 0.78
= 118N
Fapp = Fn - Fg cos theta
= 118 - mg cos 45 degrees
= 118 - (22) (9.8) (0.707)
= 118 - 152
= -34N
Could that be right?
a) Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.
b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.
For a, this is what I did:
Fapp = Fs + Fg sin theta
= Usmgcos theta + mg sin theta
= mg (Uscos theta + sin theta)
= (22kg) (9.8m/s^2)(0.78 cos 45 degrees + sin 45 degrees)
= 270N
The largest force that can be applied upward if the box is to stay at rest is 270N.
For b..i'm not totally sure but this is what I got:
Fs = Fg sin theta
= mg sin 45 degrees
= (22) (9.8) (0.707)
= 152N
Fn = Fs * Us
= 152 * 0.78
= 118N
Fapp = Fn - Fg cos theta
= 118 - mg cos 45 degrees
= 118 - (22) (9.8) (0.707)
= 118 - 152
= -34N
Could that be right?