
#1
May1910, 03:16 AM

P: 63

1. The problem statement, all variables and given/known data
a bar AB of no mass and 1m length is balanced on a support P at its midpoint. 0.6kg is hung on one end and 1kg is hung 0.3m away from the mid point on the other side. What is the upward force from support P 2. Relevant equations torque = perpendicular force * radius 3. The attempt at a solution i thought it would be simple. i figured out the force on one side was 3N and the other side was the same. So 3N + 3N both acting down is 6 N since the bar isnt moving down it must be being pushed from the support with a force of 6N. However the answer in my book is 16N i don't understand how it can possibly be this. Please show me where i went wrong. Thank you. 



#2
May1910, 03:34 AM

P: 99

If I place a 1kg box on a table, what is the normal force exerted by the table on the box? Does this force depend on how the mass is distributed within the box? Going back to your problem, would it matter if both weights were placed at the midpoint of the bar?




#3
May1910, 03:39 AM

P: 63

so your saying i just add up the two forces 10 + 6=16N ? That's the correct answer but why wouldn't i need to use force * radius?
I think it does matter if they are placed away from the midpoint 



#4
May1910, 03:42 AM

P: 63

Simple question about torque / gravity / force
Actually i think i may understand this. Because the torque on each side is equal the bar remains horizontal, so the force of each is directly down and counteracting the rotation caused by the other mass.




#5
May1910, 03:52 AM

P: 99

Well the problem as you've stated it is just asking for the upward force from the support which is touching the bar at the midpoint. Which is just the weight of the two masses correct? There is no calculation of torque needed to solve this problem. Now you're correct about the net torque on the bar depending on where the masses are placed since the equation for torque involves the radius from the pivot. So if the torque on each side wasn't balanced the bar would rotate. But as it is, they are balanced in this problem and the net torque on the bar is zero. This would also be true if the two masses were placed at the midpoint of the bar where the support P is, since the force due to their combined weight would be acting on the pivot, in which case they would not contribute to the torque on the bar and the net torque would again be zero. Either way as long as the bar doesn't fall off the pivot, which could correspond to a number of combinations of where the two masses are placed, then the normal force is just equal to the weight of the two masses; regardless of their position.




#6
May1910, 04:02 AM

P: 63

thank you for your prompt help



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