
#1
May2504, 10:53 PM

P: 5

I'm stuck on these problems!
1. Suppose a 90.0g 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 2000kg 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. 



#2
May2604, 01:03 AM

P: 15

set up the momentum balance about the zero mark of the meterstick:
90*9.8*82+120*9.8*50210*9.8*x=0 210 (=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. 



#3
May2604, 04:46 AM

Mentor
P: 40,875

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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