Asked by Amy
A block of mass 3 kg is at rest on an incline. The coefficient of state friction between the block and incline is 0.34. What is the maximum angle of the incline such that the block will not slide down the plane?
Answers
Answered by
Henry
Wb = m * g = 3kg * 9.8N/kg = 29.4 N. = Wt. of the block.
Fb = 29.4N. @ Ao. = Force of the block.
Fp = 29.4*sinA = Force parallel to the incline.
Fv = 29.4*cosA = Force perpendicular to the incline.
Fs = u*Fv = 0.34 * 29.4*cosA = 10*cosA.
Fp-Fs = m*a.
29.4*sinA-10*cosA = m*0 = 0
29.4*sinA = 10*cosA
Divide both sides by cosA:
29.4*(sinA/cosA) = 10
Replace sinA/cosA with tanA:
29.4*tanA = 10
tanA = 0.34
A = 18.8o
Fb = 29.4N. @ Ao. = Force of the block.
Fp = 29.4*sinA = Force parallel to the incline.
Fv = 29.4*cosA = Force perpendicular to the incline.
Fs = u*Fv = 0.34 * 29.4*cosA = 10*cosA.
Fp-Fs = m*a.
29.4*sinA-10*cosA = m*0 = 0
29.4*sinA = 10*cosA
Divide both sides by cosA:
29.4*(sinA/cosA) = 10
Replace sinA/cosA with tanA:
29.4*tanA = 10
tanA = 0.34
A = 18.8o
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