Calculating Upward Forces on a Rigid Beam Supported by Two Points

[Dark Visitor]

I am unsure of how to go about this problem, because whatever I tried wasn't right. So if anyone could help me by going step by step through it, showing all equations and numbers used, and then showing me the answers so I can make sure I get the same thing, I would appreciate it.

A 3.0-m-long rigid beam with a mass 150 kg is supported at each end. An 70 kg student stands 2.0 m from support 1. The board is 3 m long. (The picture I provided the link for may be helpful.)

a) How much upward force does support 1 exert on the beam?
b) How much upward force does support 2 exert on the beam?

http://session.masteringphysics.com/problemAsset/1035193/4/jfk.Figure.P08.35.jpg

 

[Andrew Mason]

I am unsure of how to go about this problem, because whatever I tried wasn't right. So if anyone could help me by going step by step through it, showing all equations and numbers used, and then showing me the answers so I can make sure I get the same thing, I would appreciate it.

A 3.0-m-long rigid beam with a mass 150 kg is supported at each end. An 70 kg student stands 2.0 m from support 1. The board is 3 m long. (The picture I provided the link for may be helpful.)

a) How much upward force does support 1 exert on the beam?
b) How much upward force does support 2 exert on the beam?

http://session.masteringphysics.com/problemAsset/1035193/4/jfk.Figure.P08.35.jpg

The upward force of the right support x the separation (3 m) is the torque applied by that support with respect to rotation about fulcrum of the left support. That torque is balanced by the sum of the downward torques created by the weights of the beam and student. Since there is no net angular acceleration, the two torques are equal in magnitude and opposite in direction.

So work out the downward torques. Think of the beam as having a centre of mass located in the middle. What torque is being applied by the beam about the right fulcrum? What torque is being applied by the student about the right fulcrum? That gives you the upward torque and, hence, the upward force on the right support.

How would you quickly determine the upward force of the left support after working out the upward force of the right?

AM

 

[Dark Visitor]

I think the torque being applied to the right fulcrum would be half of the total torque? And the student would have 2/3 the torque because he is closer to the right support.

Once I get that, the left support would be the other 1/3 of the previous answer?

 

[Andrew Mason]

I think the torque being applied to the right fulcrum would be half of the total torque? And the student would have 2/3 the torque because he is closer to the right support.

Once I get that, the left support would be the other 1/3 of the previous answer?

Use the definition of torque:

\tau = \vec F \times \vec d

|\tau |= |\vec F||\vec d|\sin\theta = mgd\sin{(90)} = mgd

where d is the distance from the point where the force is applied, to the fulcrum.

For the beam, you have to integrate \int xdm over the whole beam but that is equivalent to simply taking the mass of the beam as a point mass located at the centre of mass.

AM

 

[Dark Visitor]

So are you saying I find mgd, and then put that number over weight at the center of the board?

 

[Andrew Mason]

So are you saying I find mgd, and then put that number over weight at the center of the board?

No. You have to find the downward moments of two things with respect to one of the supports: 1. the weight of the beam located at 1.5 m from each support and 2. the weight of the student located 2 m. from the left and 1 m. from the right. The sum of those downward moments about a support is equal to the upward moment of the other support.

The downward moment of the man about the left support is mgd (in Nm) where m is the student mass and d is the distance to the left support. The downward moment of the beam is mgd where m is the mass of the beam and d is the distance of its centre of mass to the left support. Add those together. That moment is exactly equal to the upward moment of the other support.

AM

 

[CJSGrailKnigh]

Use the equillibrium equations.

\SigmaFx=0
\SigmaFy=0
\SigmaM=0 (at any point along the beam.)

Remember that when taking the moment at a point you don't have to worry about any unknown there. So take the moment about the point that eliminates as many unknowns as possible.

sorry just noticed you were using torque... moment is torque just an engineering term.