Asked by Yahaya
A 5 kg block is placed on an inclined plane raised to angle theta with the horizontal when it just started to slip down. If coefficient of friction is 0.3, calculate theta and the speed of the block up the plane if it being pull up by a force of 30 newton
Answers
Answered by
Henry
a. Wb = m*g = 5kg * 9.8N/kg = 49 N. = Wt. of block.
Fb = 49N. @ A Drees
Fp = 49*sinA = Force parallel to plane.
Fv = 49*cosA = Force perpendicular to plane or Normal.
Fs = u*Fv = 0.3*49*cosA = Force of static friction.
Fp-Fs = m*a.
49*sinA-0.3*49*cosA = m*a = m*0 = 0.
49*sinA-14.7*cosA = 0
49*sinA = 14.7*cosA
sinA = 0.3*cosA
Divide both sides by cosA:
SinA/cosA = 0.3
Replace sinA/cosA with tanA:
tanA = 0.3
A = 16.7o = The angle at which the block starts to slip.
Fb = 49N. @ A Drees
Fp = 49*sinA = Force parallel to plane.
Fv = 49*cosA = Force perpendicular to plane or Normal.
Fs = u*Fv = 0.3*49*cosA = Force of static friction.
Fp-Fs = m*a.
49*sinA-0.3*49*cosA = m*a = m*0 = 0.
49*sinA-14.7*cosA = 0
49*sinA = 14.7*cosA
sinA = 0.3*cosA
Divide both sides by cosA:
SinA/cosA = 0.3
Replace sinA/cosA with tanA:
tanA = 0.3
A = 16.7o = The angle at which the block starts to slip.
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