A non-uniform, 2.0 m long ladder weighing 400 N leans against a smooth vertical

Question

A non-uniform, 2.0 m long ladder weighing 400 N leans against a smooth vertical wall (no friction). The ladder makes an angle of 30 degree with the vertical. The coefficient of static friction at the ground is 0.3. A child with weight of 300 N can climb the ladder of 1.2 m before it starts to slide down. The distance of the center of gravity of the ladder from the contact point in the ground is The normal force coming from the smooth wall is Starting from rest, a uniform cylindrical wheel (I = M R^2) with mass M = 20 kg and radius R = 0.5 m is rotating under a constant angular acceleration of 60 rad/s^2. The magnitude of the total acceleration of a point on the rim of the wheel at t = 0.2 s is The magnitude of the angular momentum of the cylindrical wheel at t = 0.5 s is

Explanation / Answer

along vertical

Fnet = 0

N1 - W1 - w2 = 0

N1 = 400+200 = 600 N

along horizantal

f = N2

u*N1 = N2

In equilibrium net torque about the point of contace = 0

W1*l1*sin30 + w2*l2*sin30 = N2*l*cos30

(400*l1*sin30)+(200*1.2*sin30) = (0.3*600*2*cos30)

l1 = 0.967 m <<----answer

b)

N2 = (0.3*600) = 180 N

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3)

at t= 0.2 s

tangential acceleration = r*alpha = 0.5*60 = 30 m/s^2

w = alpha*t = 60*0.2 = 12 rad/s

radial acceleration = ar = r*w^2 = 0.5*12^2 = 72 m/s^2

total acceleration = a = sqrt(at^2+ar^2) = sqrt(30^2+72^2) = 78 m/s^2

b)

at t = 0.5 s

w = alpha*t = 60*0.5 = 30 rad/s

angular momentum L = I*w = M*R^2*w = 20*0.5^2*30 = 150 kg m^2/s

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