Question

our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three on the right (total mass mp=10.1 kgmp=10.1 kg). The distance from the fulcrum to the center of the pile of rocks is rp=0.360 m.rp=0.360 m. Answer the two questions below, using three significant digits. Part A: What is the value of the torque (????pτp) produced by the pile of rocks? (Enter a positive value.)

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$  

b

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

     The mass of the left rock is  $m_s = 2.25 \ kg$

     The mass of the rock on the right $m_p = 10.1 kg$

    The distance from  fulcrum to the center of the pile of rocks is  $r_p = 0.360 \ m$

   

Generally the torque produced by the pile of rock is mathematically represented as

           $\tau = m_p * g * r_p$

Substituting values

         $\tau = 10.1 * 9.8 * 0.360$                  

          $\tau = 35.63\ N \cdot m$      

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

           $\tau = m_s * g * r_s$

At equilibrium the both torque are equal

            $35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

             $r_s = \frac{35.63 }{m_s * g}$

Substituting values

            $r_s = \frac{35.63 }{2.25 * 9.8}$

            $r_s =1.62 \ m$