Center Crib Toy
August 30th 2010 Posted at Baby Toyts
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Center Crib Toy
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Challenging Torque Problems???
1.) Tracy is building a mobile to hang over her baby's crib. She hangs a 0.02 kg toy sailboat 0.01 m from the left end and a 0.015 kg toy truck 0.2 m from the right end of a bar 0.5 m long. If the lever arm itself has negligible mass, where must the support string be placed so that the arm balances?
2.) Orin and Ann, two paramedics, rush a 60 kg man from the scene of an accident to a waiting ambulance, carrying him on a uniform 3 kg stretcher held by the end. The stretcher is 2.6 m long and the man's center of mass is 1 m from Ann. How much force much Orin and Ann each exert to keep the man horizontal?
Problem 1:
Your first step is to determine how much force there is in the support string. This is arrived at by getting the summation of all the vertical forces and since the system has to be in equilibrium
Force in the support string (call this S) = 0.015 + 0.02 = 0.035
The next step is to locate where S will be along the mobile and this is done by getting the summation of all torques and again, since this system is in equilibrium, the summation of torques = 0.
Let X = distance from the right end of the bar where the support string will be placed
NOTE: All distances for this torque problem will be reckoned from the right side of the bar.
((0.015)(0.2)) - (0.035X) + ((0.02)(0.48)) = 0
Rearranging your equation,
0.035X = ((0.015)(0.2)) + ((0.02)(0.48))
and solving for X,
X = (0.003 + 0.0096)/0.035
X = 0.36 m
Answer: The support string will have to be placed 0.36 m from the right side of the bar.
Problem 2:
Let
X = portion of the load that Ann has to carry
Y = portion of the load that Orin has to carry
Since the system is in equilibrium, the summation of all vertical forces must be equal. Hence,
X + Y = 60 + 3 = 63
Since there are 2 unknowns, there needs to be one more equation so this problem can be solved. To do this, get a summation of torques and again, since this system is in equilibrium, the summation of torques should be equal to zero.
NOTE: All distances to be used in the torque formula wil be reckoned from Ann's end.
(60)(1) + (3)(1.3) - Y(2.6) = 0
Rearranging,
2.6Y = (60)(1) + (3)(1.3) = 60 + 3.9
2.6Y = 63.9
and solving for Y,
Y = 63.9/2.6 = 24.57
Since X + Y = 63.9, then
X = 63.9 - 24.57 = 38.43
Answer:
Ann's portion of the load = 38.43 kg.
Orin's portion of the load = 24.57 kg.
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