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Question

A weight is attached to a rope 42 ft long. The rope passes over a pulley 22 ft above the ground. A man takes hold of the end of the rope and walks back at 10 ft per second holding the end at a level of 6 ft above the ground. At what rate is the weight ascending when it is 6 ft above the ground?

Answer 1

PROBLEM

A weight is attached to a rope 42 ft long. The rope passes over a pulley 22 ft above the ground. A man takes hold of the end of the rope and walks back at 10 ft per second holding the end at a level of 6 ft above the ground. At what rate is the weight ascending when it is 6 ft above the ground?

SOLUTION

DIAGRAM OF ROPE, PULLEY, WEIGHT, AND MAN

PULLEY (22 FT ABOVE GROUND)

                                                     o                                    -----------------
                                                     | \                                            /\
                                                     |  \                                            | ROPE LENGTH TO RIGHT OF PULLEY
                                                     |   \ <------------------------- c (HYPOTENUSE) = 26 FT WHEN WEIGHT 6 FT
                                                     |    \                                          | ABOVE GROUND 
  LENGTH OF ROPE LEFT OF PULLEY  |     \                                         |
l = 42 - c = 16 FT WHEN WEIGHT     |      \                                        |
IS 6 FT ABOVE GROUND                  |        \                                      | <---------- a (VERTICAL LEG) = 16 FT SINCE
                                                     |         \                                    \/ MAN KEEPS ROPE 6 FT ABOVE GROUND
                                                     |          \ _o_                   -------------------
                                                     |               |
                                                    # _______ /\ ____________________

                                                     |-----------| <--------- b (HORIZONTAL LEG) = √420 WHEN WEIGHT IS 6 FT
                                                                                                                         ABOVE GROUND

CALCULATIONS AND EQUATIONS

l (LENGTH OF ROPE LEFT OF PULLEY) = 42 - c = 16 FT WHEN WEIGHT IS 6 FT ABOVE GROUND

a (VERTICAL DISTANCE OF RIGHT ROPE END TO PULLEY) = 16 FT CONSTANT SINCE MAN KEEPS ROPE END AT A
LEVEL OF 6 FT ABOVE GROUND

b (HORIZONTAL DISTANCE OF RIGHT ROPE END TO PULLEY) = √420 WHEN WEIGHT IS 6 FT ABOVE GROUND;
FOUND BY USING PYTHAGOREAN THEOREM ==> b = SQRT (c^2 - a^2) = SQRT (26^2 - 16^2) = √420

c (LENGTH OF ROPE RIGHT OF PULLEY) = 42 - l = 26 FT WHEN WEIGHT IS 6 FT ABOVE GROUND


RIGHT END OF ROPE MOVES IN HORIZONTAL DIRECTION 10 FT/S = db/dt

ASCENT OF LEFT END OF ROPE = dl/dt

SINCE l = 42 - c, AND c = SQRT (a^2 + b^2) = (a^2 + b^2)^(1/2)

AND c = 26, a = 16, and b = √420 WHEN WEIGHT IS 6 FT ABOVE GROUND

dl/dt = d/dt [42 - c] = d/dt (42) - d/dt [(16^2 + b^2)^(1/2)]

dl/dt = 0 - [(1/2)(16^2 + b^2)^(-1/2)](2)db/dt

dl/dt = - [(16^2 + b^2)^(-1/2)]db/dt

WE CAN SUBSTITUTE c^2 FOR (16^2 + b^2); THEREFORE,

dl/dt = - [(c^2)^(-1/2)]db/dt

SUBSTITUTING VALUES FOR c^2 AND db/dt

dl/dt = - [(676)^(-1/2)](10)

dl/dt = - [1/26](10)

dl/dt = - 10/26 = - 5/13 FT/S = 0.384615 (REPEATING) FT/S <--- NEGATIVE SIGN MEANS LEFT LENGTH OF ROPE SHORTENING

THEREFORE, RATE OF ASCENT OF WEIGHT AT LEFT END OF ROPE = 5/13 FT/S (~0.38 FT/S)