A gymnast with mass m 1 = 45 kg is on a balance beam that sits on (but is not at

Question

A gymnast with mass m1 = 45 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 114 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.

1)

What is the force the left support exerts on the beam?
N

2)

What is the force the right support exerts on the beam?
N

3)

How much extra mass could the gymnast hold before the beam begins to tip?
kg

4)

Now the gymnast (not holding any additional mass) walks directly above the right support.

What is the force the left support exerts on the beam?
N

5)

What is the force the right support exerts on the beam?
N

6)

At what location does the gymnast need to stand to maximize the force on the right support?

at the center of the beam

at the right support

at the right edge of the beam

Explanation / Answer

Draw a diagram of your setup.
Call the upwards supporting force from the left support on the beam, L.
Call the upwards supporting force from the right support on the beam, R.
Weight of m1 = 45 x 9.8 = 441N downwards from left edge.
Weight of beam = 114 x 9.8 = 1117.2N downwards from centre of gravity of beam (middle).

Make sure you mark L, R and weights with arrows on your diagram.
_____________________________

From diagram you should see:
Distance of gymnast from left support = 5/3 m. Put this on diagram.
Distance of beam's centre of gravity from left support = 5/2 - 5/3 = 5/6 m. Put this on diagram.
Distance of L from left support = 0
Distance of R from left support = (2/3) x 5 = 10/3 m. Put this on diagram.

Take moments (torques) about the left support:
Moment produced by gymnast = 441 x (5/3) anticlockwise
Moment produced by weight of beam = 1117.2 x 5/6 clockwise
Moment produced by L = 0 because L acts through the left support
Moment produced by R = (10/3)R anticlockwise

1)The total moment is zero as the beam is in equilibrium.
clockwise moments = anticlockwise moments
1117.2 x (5/6) = 441 x (5/3) + (10/3)R (equation 1)
931 = 735 + 3.33R
R = 58.85N ANSWER
__________________________________

2)Since the beam is in equilibrium, the total upwards forces (L+R) equal the total downwards forces (the 2 weights)
=>L + R = 1038.8 + 421.4

=>L=1460.2-58.85

=>L= 1401.35 N ANSWER

3)To find the extra mass (M). Its weight = Mg = 9.8M.
Weight of gymnast plus M = (441 + 9.8M)
When just about to tip, R=0

So equation 1 becomes:
(1117.2 + 9.8M) x (5/6) = 441 x (5/3) + (10/3)x0
which you can solve for M

M= 24.01 KG

4) (114 x 9.8) = 1117.2N.
(45 x 9.8) = 441N.
It is not important where the supports are placed, as without the gymnast on the beam, the weight is evenly distributed, so long as both supports are the same distance from the ends.
Weight on each support = (1117.2/2) = 558.6N.
It remains 558.6N.
5) (558.6 + 441) = 999.6N.

6) at the right edge of the beam

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