Solved:
Answered by AI, Verified by Human Experts
The solid homogeneous cylinder is released from rest on the ramp. If θ= 40° , µs= 0.30 and µk= 0.20. Determine the magnitudes of the acceleration of the mass (W= 8lb) center G and the friction force exerted by the ramp on the cylinder.
The solid homogeneous cylinder is released from rest on the ramp. If θ= 40° , µs= 0.30 and µk= 0.20. Determine the magnitudes of the acceleration of the mass (W= 8lb) center G and the friction force exerted by the ramp on the cylinder.
The magnitudes of theaccelerationand thefriction forceare;Acceleration = 13.8 ft/s²Friction Force = 1.714 lbThe image of the solid homogeneous cylinder is missing and so i have attached it.From the image we see that;Weight; W = 8 lbRadius; r = 6 in = 0.5 ftWe are given;Angle of incline; θ = 40°Coefficient of static friction; µ_s = 0.30coefficient of kinetic friction; µ_k = 0.20We know that formula for weight is; W = mgThus; m = W/gwhere g isacceleration due to gravity= 32.2 ft/s²m = 8/32.2mass; m = 0.2484 lb.s²/ftNow, to get theaccelerationalong thex-axis, we will use the formula;a = rαwhere α isangular acceleration. Thus;a = 0.5αα = 2a ----- (eq 1)Now,resolving forcesalong thex-directiongives;W*sinθ - F = maPlugging in the relevant values;8*sin 40 - F = 0.2484aF = 8*sin 40 - 0.2484a -----(eq 2)Now,moment of inertiaof the cylinder along thez-axisis gotten from;I = ¹/₂mr²I = ¹/₂ × 0.2484 × 0.5²I = 0.03105 lb.ft/s²Takingequilibrium of momentswe have;F*r = I*αThus;(8*sin 40 - 0.2484a)0.5 = 0.03105α⇒ 2.57115 - 0.1242a = 0.03105α⇒ 0.03105α + 0.1242a = 2.57115From eq 1, α = 2a. Thus;0.03105(2a) + 0.1242a = 2.571150.1863a = 2.57115a = 2.57115/0.1863a = 13.8 ft/s²Formula for thefriction forceexerted by the ramp on the cylinder is;F = 8*sin 40 - 0.2484aF = 5.1423 - 0.2484(13.8)F = 1.714 lbRead more aboutcylinder moment of inertiaat;brainly.com/question/7020147...