|May25-04, 10:53 PM||#1|
I'm stuck on these problems!
1. Suppose a 90.0-g mass is placed at the 82.0 cm mark of a 120.0 g uniform meterstick. At what mark will the meterstick balance?
For this problem, I did TCCW=TCW. I think the TCW is (90 g)*(9.8 m/s^2)*(32 cm). For TCCW, I am not sure because I do not know where the pivot point is.
2. A 10,000 kg bridge of length 10 m is supported at both ends. If a 2000-kg car is parked on the bridge 3.0 m from the left support, what are the supporting forces at the left and right ends?
I drew a diagram for this problem, but I do not know how to calculate the supporting forces.
|May26-04, 01:03 AM||#2|
set up the momentum balance about the zero mark of the meterstick:
210 (=90+120)*9.8 being the reaction force at the pivotpoint
x = the distance of the pivot point from the zero mark
work out x and you have the answer for question 1.
|May26-04, 04:46 AM||#3|
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Now apply the conditions for equilibrium:
(1) The forces must balance
(2) The torques about any point must balance
Give it a shot.
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