Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The torque produced by the pile of rocks is [tex]\tau = 35.63\ N \cdot m[/tex]
b
The distance of the single for equilibrium to occur is [tex]r_s =1.62 \ m[/tex]
Explanation:
From the question we are told that
The mass of the left rock is [tex]m_s = 2.25 \ kg[/tex]
The mass of the rock on the right [tex]m_p = 10.1 kg[/tex]
The distance from fulcrum to the center of the pile of rocks is [tex]r_p = 0.360 \ m[/tex]
Generally the torque produced by the pile of rock is mathematically represented as
[tex]\tau = m_p * g * r_p[/tex]
Substituting values
[tex]\tau = 10.1 * 9.8 * 0.360[/tex]
[tex]\tau = 35.63\ N \cdot m[/tex]
Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows
The torque due to the single rock is
[tex]\tau = m_s * g * r_s[/tex]
At equilibrium the both torque are equal
[tex]35.63 = m_s * r_s * g[/tex]
Making [tex]r_s[/tex] the subject of the formula
[tex]r_s = \frac{35.63 }{m_s * g}[/tex]
Substituting values
[tex]r_s = \frac{35.63 }{2.25 * 9.8}[/tex]
[tex]r_s =1.62 \ m[/tex]
