1) A barrel is rolled up an incline 20 ft along its length and 6 ft in vertical

ID: 1857693 • Letter: 1

Question

1) A barrel is rolled up an incline 20 ft along its length and 6 ft in vertical height by means of a rope that is

fastened to the top of the incline, feeds along the incline, then feeds under, around and over the barrel, and

returns to the top, parallel to the incline and above it. Sketch first.

a. neglecting friction, to roll a barrel that weighs 500lbf , what force must a man pull with on the return

rope?

b. what is the theoretical mechanical advantage obtained by the man who pulls on the return rope to roll

the heavy barrel up the incline? (Use m* as the ratio of output force to input force.)

c. what is the ratio of the displacements of barrel moving up with respect to the displacement of the rope

being pulled? (what is the gear ratio g?)

d. with friction consider the rope is 95% efficient and the incline is 90% efficient, what actual force must a

man pull with on the return rope to roll a barrel that weighs 500lbf?

Incline angle theta = asin (6/20) = 17.46 deg

Torque = I*alpha

F*r + Fric*r = (1/2*mr^2)*(a/r)

F = 1/2*ma - Fric

Forces balance: F - Fric - mg Sin theta = ma

1/2*ma - 2*Fric - mg Sin theta = ma

2*Fric = -1/2*ma - mg Sin theta

Neglecting Fric, we get mg Sin theta = 1/2*ma

or a = 2g sin theta

F = 1/2*ma = 1/2*m*2g sin theta = mg sin theta = 500 sin 17.46 = 150 lbf

Mech. advantage = Output force / input force = ma / F = 2

s = 1/

For pure rolling , point of application of force moves twice as much as the centroid of mass.

Hence, g = 1/2 = 0.5.

Actual force = Ideal force / Efficiencies = 150 / (0.9*0.95) = 175.4 lbf

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