To carry an m = 200 pound cylindrical weight, two workers lift on the ends of sh

Question

To carry an m = 200 pound cylindrical weight, two workers lift on the ends of short ropes tied to an eyelet on the top center of the cylinder. One rope makes a 20° angle away from the vertical and the other makes a 30° angle (see figure). (Let u be the right rope and v be the left rope. Round your answers to two decimal places.)

a) Find each rope's tension if the resultant force is vertical.( That is find it for v and u).

b) Find the vertical component of each worker's force. ( also for u and v)

Explanation / Answer

Let A be the 20 degree side
Let B be the 30 degree side
F(Ahorz) = F(Bhorz) (equal and opposite)
(1) T(A) sin20 + T(B) sin30 = 100
(2) T(A) cos20 = T(B) cos30
(1) T(A) = (100/sin20) - (T(B) cos30/cos20)
(100/sin20) - (T(B) cos30/cos20) = T(B)cos30/cos20
T(B) = (100/sin20)/[(cos30/cos20) + (sin30/sin20)]
T(B) = 122.668 T

T(A)= T(B) cos30 /cos20

T(A)=113.05 T
F(Bvert) = 122.668 sin30 = 61.334 lb.
F(Avert) = 100 - 61.334 = 38.666 lb.

Ans a)Tension in rope A= 113.05 T

Tension in rope B= 122.668 T

b)

Vertical force in A= 38.666 lb

Vertical force in B= 61.334 lb

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