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A gymnast with mass 56.5 kg stands on the end of a uniform balance beam as shown in the figure below. The beam is 5.00 m long and has a mass of 250 kg (excluding the mass of the two supports). Each support is 0.530 m from its end of the beam. In unit-vector notation, what are the forces on the beam due to support 1 and support 2?

F=ma

Torque=I*alpha

and where there is no acceleration, the sums are equal to 0.

Setting up the forces as the following:

F1 = support 1

F2 = support 2

F3 = gravitational force in center of beam

F4 = force of the gymnast at the end

I used Newton 2 for the sum of the forces in equilibrium (acceleration = 0)...

F1+F2+F3+F4=ma=0 where F3 = (250kg*g), F4=(56.5kg*g)

F1+F2+306.5g=0

1.) F2=-F1-306.5g

Then Newton II for sum of the torques.... Where L is the length of the beam, and point I will rotate is located at the point where the gymnast is, so I can simplify the calculations.

2.)F1*4.47m +F3*2.5m+F2*0.053m=0

Next, I plugged 1( solved F2) into (2)

Simplifying, I get F1=1150.52N

Please help, and thank you in advance for you time and generosity.

## Homework Equations

F=ma

Torque=I*alpha

and where there is no acceleration, the sums are equal to 0.

Setting up the forces as the following:

F1 = support 1

F2 = support 2

F3 = gravitational force in center of beam

F4 = force of the gymnast at the end

I used Newton 2 for the sum of the forces in equilibrium (acceleration = 0)...

F1+F2+F3+F4=ma=0 where F3 = (250kg*g), F4=(56.5kg*g)

F1+F2+306.5g=0

1.) F2=-F1-306.5g

Then Newton II for sum of the torques.... Where L is the length of the beam, and point I will rotate is located at the point where the gymnast is, so I can simplify the calculations.

2.)F1*4.47m +F3*2.5m+F2*0.053m=0

Next, I plugged 1( solved F2) into (2)

Simplifying, I get F1=1150.52N

Please help, and thank you in advance for you time and generosity.

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