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ID: 1567513 • Letter: I
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Problem a2 (7 points) A horizontally oriented uniform rod has a weight of w 850 N and a length L 3.5 m. The rod is suspended by two vertical wires. The wires are connected Az to the rod at distances Axi 0.45 m and Axa 0.68 m from and right ends of the rod as shown in the figure. A very small object of mass m 35 kg placed at the left end of the rod. If the system is in static equnibrium find: a) the tension (in N) in the left wire (1 point); (b) the tension (in N) in the right wire (1 point) (c) If you need to have tensions in each wire to be the same by what distance (measured from the left end of the rod in m) do you need to move the mass m to the right? (2 points)Explanation / Answer
given
L = 3.5 m
a)
As the system is in equilibrium, net force and net torque acting on the rod must be zero.
let TL and TR are the tensions in the left and right strings.
a) Apply net torque about right string = 0
TL*(L - (delta_x1 + delta_x2)) - (L - delta_x2)*m*g - (L/2 - delta_x2)*W = 0
TL = ((L - delta_x2)*m*g + (L/2 - delta_x2)*W)/(L - (delta_x1 + delta_x2))
= ((3.5 - 0.68)*35*9.8 + (3.5/2 - 0.68)*850)/(3.5 - (0.68 + 0.45))
= 792 N <<<<<<<<<<------------------Answer
b) now apply, Fnety = 0
TL + TR - W - m*g = 0
TR = W + m*g - TL
= 850 + 35*9.8 - 792
= 401 N <<<<<<<<<<------------------Answer
c) let x is the distance where m is supposed to be placed from left end.
let TL = TR
use, Fnety = 0
TL + TR - W - m*g = 0
2*TL = W + m*g
TL =(W + m*g)/2
= (850 + 35*9.8)/2
= 596.5 N
Apply net torque about right string = 0
TL*(L - delta_x2)) - (L - delta_x2 - x)*m*g - (L/2 - delta_x2)*W = 0
596.5*(3.5 - 0.68) - (3.5 - 0.68 - x)*35*9.8 - (3.5/2 - 0.68)*850 = 0
==> x = 0.567 m <<<<<<<<<<------------------Answer
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