In summary: In equilibrium situations, there are usually only two forces acting on a body: the weight of the object and the force of gravity. Moment is just a measure of how much the weight of the object affects the rotation of the object. If you have two supports, and you know the mass of the object on the supports and the weight of the object at the pivot, then you can solve for the moments about the supports.In summary, a beam of 120 kg is resting on 2 supports, each of which exerts 588 N of force. The left support has a greater force than the right support.
#36
jjd101
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UPDATE: I just got an email from the teacher and the mass of the second student is 70kg, i calculated this with my understanding and got 1274N on right support and 1568N on left support if i am wrong please correct me
Plug them into sum of forces = zero and sum of torques = zero and see if they work.
#38
jjd101
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the way i evaluated they even out. i got torque on first support from student one is 980 and torque on support two from student two is 656, both supports share weight of bar evenly which is 588 each
#39
netgypsy
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No the supports do not necessarily share the weight evenly. You have to use force upward on the support on the left as a different variable from force upward on the support on the right. Your answers prove they are in fact different. It's two equations with two unknowns. Use an axis of rotation at one of the upward supports for your torque equation.
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#40
jjd101
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even though they are both placed in the same spot relevant to the bar?(2 feet from the edge) ?? how do i solve this then?
#41
netgypsy
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it takes two equations. sum the forces and set equal to zero and sum the torques and set equal to zero.
Or Add the up forces and set them equal to the down forces and for your second equation add the clockwise torques and set them equal to the counterclockwise torques.
#42
jjd101
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Support 1 Force + Support 2 Force = Torque on support 1 + torque on support 2 + weight of beam? how do i deal with clockwise torques i am unfamiliar with this
#43
netgypsy
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No do only forces for one equation - add the up forces and set them equal to the down forces. You'll have a variable for the left upward support force and another one for the right upward support force.
then do only torques for the second equation
#44
jjd101
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force of student one = 9.8*50 = 490 force of student 2 = 9.8*70 = 686 force of beam = 9.8*120= 1176 Total down force =2352 = Support 1 force + support 2 force?
#45
netgypsy
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correct so far so good.
now you need to put the axis of rotation or your "pretend" point around which this plank will rotate at one of the two unknown upward forces and then add the clockwise torques and set them equal to the counter clockwise torques. Remember you have the equation listed torque = force times perpendicular distance from the force to the axis of rotation.
#46
jjd101
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so counter clockwise from first support, 490* 2
counter clockwise from second support 490* 10
clockwise from first support 686*9
clockwise from second support 686*1
these are not even though
#47
netgypsy
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The first one on the left should be the guy at the end. He is two meters from the axis of rotation and he is counter clockwise. so 490N (2m) is correct for that one. The nextcounterclockwise is the upward force from the second support which is F right support (8m not 10meters) Your axis of rotation is at the left support which is 8m from the right support.
Clockwise you have the weight of the board downward times its distance from the left support. You don't have this one at all. and the last torque clockwise is from the guy on the right who is 1 meter from the right end so that one is the 686N(9m) so you are missing the clockwise torque caused by the weight of the board in the middle.
Once you find it, add the clockwise and set equal to the counterwise torques and solve for your only unknown. Then plug it back into the first equation you found
#48
jjd101
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clockwise torque of the board is 1176* 4m = 4704?
#49
netgypsy
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Very good. So you see when you sum the clockwise and counterclockwise torques and set them equal to each other you have only one unknown, the upward force on the right side which produces a counterclockwise torque.
than from equation 1 you wrote summing just the forces, you can find your other upward force on the left side.
#50
jjd101
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clockwise = counter clockwise
clockwise = 4704+686*9 = 10875
counterclockwise = 490*2 + Fof2ndSupport*8
solving for this 2nd support is a force of 1237N therefore the force of the first support is 2352-1237 = 1115N ?
#51
netgypsy
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You're doing it correctly. I haven't checked your math. Good job.