Wasseem92
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A poorly constructed see-saw has a fulcrum 2/3 of the way along its length. a) If the see-saw weighs 30 kg, where would a 20 kg child have to sit in order to balance the see-saw? b) What is the least mass that a child must have in order to balance the see-saw?
well i tried solving this question and my teacher gave this as a solution:
The center of gravity of the see-saw is assumed to be in the center, assuming that it is uniform. The force diagram is therefore as shown below. mss = 30 kg is the mass of the see-saw, mc = 20 kg is the mass of the child, and L is the length of the see-saw.
Figure 8.3: Problem 8.1
In order to find the distance x from the child to the fulcrum we can do a torque balance about the fulcrum:
sum of torque = (L/6)mssg - xmcg = 0
so that
x = (L*mss)/(6*mc) = = (1/4) L.
My question is: why did he start of with a distance of (L/6). Please i need help, this really baffles me.
Thank you.
well i tried solving this question and my teacher gave this as a solution:
The center of gravity of the see-saw is assumed to be in the center, assuming that it is uniform. The force diagram is therefore as shown below. mss = 30 kg is the mass of the see-saw, mc = 20 kg is the mass of the child, and L is the length of the see-saw.
Figure 8.3: Problem 8.1
In order to find the distance x from the child to the fulcrum we can do a torque balance about the fulcrum:
sum of torque = (L/6)mssg - xmcg = 0
so that
x = (L*mss)/(6*mc) = = (1/4) L.
My question is: why did he start of with a distance of (L/6). Please i need help, this really baffles me.
Thank you.
