A gymnast with mass m1 = 44 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 109 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?

a. at the center of the beam
b. at the right support
c. at the right edge of the beam

Newton's second law for linear and rotational motion allows to find the results for the questions about the strength of the support to the beam for various positions of the gymnast are:

1) Left support force is: R₁ = 1447.95 N

2) Right support force is: R₂ = 51.45 N

3) Extra mass before losing balance is: M = 10.5 kg

4) Gymnast on support, strength of support is: R₁ = 534.1 N

5) Gymnast on support, force of the left support is: R₂ = 965.3 N

6) Correct result is the answer:

c) The right edge of the beam

Given parameters

Mass of the gymnast m1 = 44 kg

Beam mass m² = 109 kg

Beam length L = 5 m

Supports ⅓ from each end.

Gymnast on the far left

To find

1) The force of the left bracket.

2) The strength of the right support.

3) Extra mass before bending over.

4) The gymnast stands on the right support, what is the strength of the left support

5) Force of the lleft support

6) Where you should stop to maximize strength.

Newton's second law for linear and rotational motion explains the linear and rotational motion of bodies, in the special case that the linear and rotational accelerations are zero, it is called equilibrium conditions;

Linear ∑ F = 0

rotational ∑ τ = 0

Where F is the force tau the troque.

The torque is defined by the vector product of the force by the position vector,

τ = F x r

The modulus of this expression is

τ = F r sin θ

A free body diagram is a diagram of the forces without the details of the bodies, in the attached we can see the free body diagram of the system.

Part 1 and 2

Let's apply the equilibrium conditions.

Linear [tex]R_1 +R_2 - w_g -w_b = 0[/tex]

For the rotational equilibrium condition we will assume positive counterclockwise rotations, the axis of rotation is in support on the left.

[tex]W_g \frac{L}{3} - W_b ( \frac{L}{2} - \frac{L}{3} ) + R_2 \frac{2L}{3} =0 \\W_g - W_b \frac{1}{6} + \frac{2}{3} R_2 =[/tex]

Let's look for the reaction of the support of the right (R₂)

[tex]R_2 = \frac{3}{2} ( - \frac{W_g}{3} + \frac{W_b}{6} = - \frac{W_g}{2} + \frac{W_b}{4}[/tex]

Let's calculate.

R₂ = (- 44/2 + 105/4) 9.8

R₂ = 51.45 N

we look for the value of the force R₁ left support.

R₁ + R₂ - W_g -W_b = 0

R₁ = -R₂ + W_g + W_b = -R₂ + (m_g + m_b) g

Let's calculate.

R₁ = -51.45 + (44 + 109) 9.8

R₁ = 1447.95 N

Part 3

The gymnast takes more extra mass, therefore the reaction in the right support (R₂) decreases at the point just before moving is worth zero

R₂ = 0

Let's write the extortion for the torque

[tex](m_g + M ) g \frac{L}{3} - m_b g \frac{L}{6} = 0 \\m_g + M = \frac{m_b}{2} \\M = \frac{m_b}{2} - m_g[/tex]

M = [tex]\frac{109}{2} - 44[/tex]

M = 10.5 kg

Part 4

For this part, the turning point must be placed on the right support, the distance of the gymnast is zero for being on the support

The equilibrium condition is:

[tex]- \frac{R_1 \ L}{3} + \frac{W_b \ L}{6} = 0 \\ R_1 = \frac{W_b}{2}[/tex]

R1 = 109 9.8 / 2

R1 = 534.1 N

Part 5

We use the linear equilibrium condition.

[tex]R_1 + R_2 - W_g -W_b = 0 \\R_2 = W_g + W_b - R_1 = ( m_g + m_b) g - R_1[/tex]

R2 = (109 +44) 9.8 - 534.1

R2 = 965.3 N

Part 6

When we compare the results of parts 1 and 5 we see that the maximum reaction force of the beam occurs when the gymnast is at the end of the bar, therefore the correct result is:

c. The right edge of the beam

In conclusion, using Newton's second law for linear and rotational motion, we can find the results for the questions about the strength of the support to the beam for various positions of the gymnast are: