A gymnast with mass m 1 = 42 kg is on a balance beam that sits on (but is not at

Question

A gymnast with mass m1 = 42 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 112 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

PLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS

Explanation / Answer

When teh gymnast is standing on the left end,

42*(9.8) + 112(9.8) = R1 + R2 ( where R1 is the normal reaction on left support, R2 is right support)

R1+R2= 1509.2

rotational equation about center of beam ( in the clockwise)

R1(5/2-5/3) = R2 ( 5/2-5/3)) + 42(9.8) ( 5/2)

(R1- R2= 1235.1 aprx

1509.2 - 2R2= 1235.1 aprx

R2= 136.95 N ( force the right support exerts on the beam)

R1= 1382.05 N (  force the lieft support exerts on the beam)

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

let's use the left support as pivot,

M = mass of gymnast and extra mass

m2= mass of beam centered at L/2

Mg( L/3) = m2g( L/2-L/3)

M(1/3) = m2( 1/6)

M= m2/2 = 112/2= 56 kg

extra mass = 56-42 = 14 kg

d) now the gymnast is on right support

R1+R2= 1509.2

taking the moment at center of beam

R1( L/2-L/3) + Mg ( L/6) = R2( L/6)

R1+Mg = R2

R1 + R1 + Mg =  1509.2

2R1+ 42(9.8) = 1509.2

R1= 548.8 N

R2= 960.4 N

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