(a) What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure 9.35 in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge?

Solution 15PE Step-by-step solution Step 1 o f 10 From the first condition of equilibrium net force acting on system is zero. The first condition for equilibrium gives, From the second condition of equilibrium net torque acting on system is zero. The second condition for equilibrium gives, Step 2 o f 10 The three forces acted on the entire sandwich board system are w eight of the sign w, acted down at the center of mass of the system, normal reaction force on right and left leg respectively acted up at the ground for each of the legs. The tension and the hinge exert internal forces, and therefore cancel when considering the entire sandwich board. From the first condition for equilibrium, By symmetry, the normal forces are equal. So, Here, m is the mass of the sign board and g is the acceleration due to gravity. Substitute 8.00 kg for m and for g . Step 3 o f 10 The tension on each leg is 39.2 N. But the total tension is equal to weight entire sign board 78.4 N.