Figure 6.2 Force diagram with masses at each end of a pivoted bar of length (L+1) m.
-In Fig. 6.2, the normal force N is 20 N and masses m_A and m_B, where m_B = 2m_A, hang in equilibrium at the ends of a uniform bar of mass m_bar = m_A/2 and length (L+1) m, exerting loads w_A and w_B, respectively. What are the values of m_bar and L that give equilibrium?

Question

Figure 6.2

Force diagram with masses at each end of a pivoted bar of length (L+1) m.
-In Fig. 6.2, the normal force N is 20 N and masses m_A and m_B, where m_B = 2m_A, hang in equilibrium at the ends of a uniform bar of mass m_bar = m_A/2 and length (L+1) m, exerting loads w_A and w_B, respectively. What are the values of m_bar and L that give equilibrium?

Quizplus · Accepted Answer

The normal force N equals the total weight of the system. Given N = 20 N, and knowing the weights are w_A = m_A * g, w_B = 2m_A * g, and w_bar = (m_A/2) * g, we solve for m_A. The equilibrium condition requires the torques about the pivot to balance, leading to L = 5/7 m. Solving for m_bar = m_A/2 gives m_bar = 0.29 kg.

Figure 62 Force Diagram with Masses at Each End of of a Pivoted