A load of concrete M hangs from a rope which passes over a pulley B. ( Fig. 3.)

Question

A load of concrete M hangs from a rope which passes over a pulley B. ( Fig. 3.) A construction worker at C pulls the rope as he walks away at 5 feet per second. The level of the pulley is 10 feet higher than his hand. At what rate is the load rising if the length BC is (a) 15 feet? (b) 100 feet?

A load of concrete M hangs from a rope which passes over a pulley B. ( Fig. 3.) A construction worker at C pulls the rope as he walks away at 5 feet per second. The level of the pulley is 10 feet higher than his hand. At what rate is the load rising if the length BC is (a) 15 feet? (b) 100 feet?

Explanation / Answer

Rate of load rising = rate of change of BC because the length of rope is constant.

Let the vertex of right triangle formed be A.

So, ABC is a right triangle with right angle at A.

AC = 10ft (given)

a) BC = 15ft

We know that AB^2 + AC^2 = BC^2

So, AB = 5*sqrt(5)

Differentiating the above equation wrt to time, we get

2.AB.d(AB)/dt = 2.BC.d(BC)/dt

d(AB)/dt = speed of man = 5ft/s

=> So, d(BC)/dt = 5*sqrt(5)*5/15 = 5*sqrt(5)/3

So, Rate of load rising = rate of change of BC = 5*sqrt(5)/3 ft/s

b) BC =100ft

=> by same method as above, AB = 30*sqrt(11)

We know that AB^2 + AC^2 = BC^2

Differentiating the above equation wrt to time, we get

2.AB.d(AB)/dt = 2.BC.d(BC)/dt

d(AB)/dt = speed of man = 5ft/s

=> So, d(BC)/dt = 30*sqrt(11)*5/100 = 3*sqrt(11)/2

So, Rate of load rising = rate of change of BC = 3*sqrt(11)/2 ft/s

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