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A box weighing 20 N, resting on a ramp, is kept at equilibrium by a 4N force at an angle of 20 degrees to the ramp, together wi...Asked by Connie
A box weighing 20N, resting on a ramp, is kept at equilibrium by a 4-N force at an angle of 20 to the ramp, together with a frictional force of 5N, parallel to the surface of the ramp. Determine the angle of elevation of the ramp.
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Answered by
Henry
Wt. of box = 20 N.
Fp1 = 4*Cos20 = 3.76 N. = Force acting Parallel to ramp(upward).
Fp2 = 20*sin A = Force acting parallel to ramp(downward).
Fp2-Fs-Fp1 = 0.
20*sin A-5-3.76 = 0. Equilibrium.
A = ?.
9
20*sin A-5-3.76 = 0.
20*sin A = 8.76,
A = 10.8o. = Angle of elevation.
Fp1 = 4*Cos20 = 3.76 N. = Force acting Parallel to ramp(upward).
Fp2 = 20*sin A = Force acting parallel to ramp(downward).
Fp2-Fs-Fp1 = 0.
20*sin A-5-3.76 = 0. Equilibrium.
A = ?.
9
20*sin A-5-3.76 = 0.
20*sin A = 8.76,
A = 10.8o. = Angle of elevation.
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