A 51-kg box is suspended from the right end of a horizontal rod that has very sm
ID: 1471497 • Letter: A
Question
A 51-kg box is suspended from the right end of a horizontal rod that has very small inertia. The left end of the rod is affixed to a wall by a pin. A wire connects the right end of the rod to the wall directly above the pin, making an angle of42 with the rod.
Calculate the tension in the wire.
Determine the magnitude of the reaction force the pivot exerts on the rod.
Calculate the tension in the wire for a 10.2-kg rod.
Determine the magnitude of the reaction force the pivot exerts on the rod for a 10.2-kg rod.
Explanation / Answer
(a)
T sin theta L = mgL
T = mg/ sin theta = 51 ( 9.8)/ sin 42 = 746.93 N
(b)
the horizontal force is
Fx = T cos 42 = 746.93 N cos 42 = 555.08 N
the vertical fore is
Fy = mg - T sin 42 = 0
the net force is
F = root Fx^2 + Fy^2 = root ( 555.08)^2 = 555.08 N
(c)
take momnet about pivot point
L T sin 42 = 51 ( 9.8) L + 10.2 * g ( L/2)
T = 821.63 N
(d)
Fx = T cos 42 = 821.63 N cos 42 = 610.59 N
Fy = 51(9.8) ( 10.2 ) 9.8) - T sin 42 = 51(9.8)+ ( 10.2 ) 9.8) - 821.63 N sin 42=49.99 N
F = root Fx^2 + Fy^2 = root 610.59^2 + ( 49.99)^2 = 612.63 N
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