(B) the Net Force on the Board Must Be Zero. Therefore

9. The length of the board is and the pedestals are apart. The diver's weight is We assume the direction of the force due to the left pedestal to be down and that due to the right pedestal to be up.

(a) We calculate the torque about the right pedestal, with the diver's weight taken to produce a positive torque:

This leads to

Since the result is positive, we were correct in assuming that this force was downward.

(b) The net force on the board must be zero. Therefore,

Since the result is positive, we were correct in assuming that this force was upward.

(c) The left pedestal produces a downward force on the board, so the force on the left pedestal is upward. It is being stretched.

(d) The right pedestal produces an upward force on the board, so the force on the right pedestal is downward. It is being compressed.

29. The forces acting on the plank are shown in the diagram below.

The force due to the roller is F, which is normal to the plank since it is frictionless. The plank's weight is W, and the ground exerts a normal force N and a frictional force f. The horizontal components of force must sum to zero, so The vertical components of force must sum to zero, so Combining these two equations leads to The torque calculated about the point where the plank touches the roller (with clockwise positive) is given by

We note that Therefore, the normal force at the floor is (since )

If we combine this result with we find

which leads to

where we have used the trigonometric identity in the final step. We can then rewrite the normal force as

where we have used the trigonometric identity in the final step. When the plank slips. At the frictional force is at its maximum value This allows us to solve for the coefficient of friction:

39. (a) Let FA and FB be the forces exerted by the wires on the log and let m be the mass of the log. The area of each wire (of radius r) is The Young's modulus of steel is Wire A had an original length and stretched by , while wire B had an original length and stretched by . We are told that wire A was initially smaller than wire B by so We are told that after stretching, the wires are the same length, so The stretching of each wire is given by Eq. 13-20, so

Since the log is in equilibrium Combining these two equations leads to We solve for the force due to wire A:

(b) From the condition we obtain