Torque You and a friend of equal strength want to carry a 200 lb. Object situate

Question

Torque You and a friend of equal strength want to carry a 200 lb. Object situated on a 30 lb. board as shown: The length of the board is 7 ft, and the center of the object is 1.5 ft from the left end of the board. You have another 150 lb friend who sits on the right end of the board. If you are carrying the board at the left end, and your friend carries an equal weight, How much force must you exert on the board to carry it? How far from you should your friend be when he is carrying the board with you?

Explanation / Answer

Acceleration due to gravity, g = 9.8 m/s2 = 32.15 ft/s2

a)
1 lb = 0.454 kg
Total downward force = (200 + 30 + 150) x 0.454 x g = 1690.696 N
Since you and your friend exerts the same upward force.
2F = 1690.696 N
F = 845.35 N

b)
200 lb is at a distance of 1.5 ft from left end, 30 lb is at a distance of 3.5 ft from the left end ( since it is the center of mass), 150 lb is at a distance of 7 ft from the left end and your friend is at r feet from the left end.
Torque exerted by the masses with respect to the left end
= 200 g x 1.5 + 30 g x 3.5 + 150 g x 7 ...(1)
Torque exerted by the friend = Fr (2)
F in terms of g and lb = 190 g
Since it is in static equilibrium,
(1) = (2)
200 g x 1.5 + 30g x 3.5 + 150g x 7 = 190g x r
r = 7.65 ft

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