In the figure below, two 2 kg masses and a 4 kg mass are suspended from either e
ID: 1617082 • Letter: I
Question
In the figure below, two 2 kg masses and a 4 kg mass are suspended from either end of a 1 meter long bar (assume the bar has no mass). You want to tie a rope (indicated by the dotted line) somewhere along the bar in order to hold up the bar and keep it balanced. The distance between the two smaller masses is 0.2 m.
(a) Around the pivot point of the 4 kg mass, what is the magnitude of the torque exerted by the 2 kg mass at the far left?
(b) Around the pivot point of the 4 kg mass, what is the magnitude of the torque exerted by the other 2 kg mass?
(c) What is the tension in the rope? (i.e., what is the force the rope must exert in order to hold the whole system up?)
(d) What is the distance of the rope from the 4 kg mass? (Hint: use the 4 kg mass as your pivot point)
2kg 2kg 4 kgExplanation / Answer
Around the pivot point of the 4 kg mass,
the magnitude of the torque exerted
by the 2 kg mass at the far left = m*g*L = 2*9.8*1 = 19.6 Nm
(b)
Around the pivot point of the 4 kg mass,
the magnitude of the torque exerted
by the other 2 kg mass = m*g*(L-0.2) = 2*9.8*(1-0.2) = 15.68 Nm
(c)
along vertical
Fnet = 0
T - mg - mg - Mg = 0
T = (m + m+ M)*g
T = (2 + 2 + 4 ) *g
T = 8*9.8 = 78.4 N
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(d)
In equilibrium net torque about 4 kg = 0
torque due to tension = -T*d = -78.4 *d
net torque = 0
19.6 + 15.68 - 78.4*d = 0
d = 0.45 m from 4 kg mass <<<------answer