Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 408 N walk on the overhanging part of the plank before it just begins to tip?

To determine the maximum distance a person can walk on the overhanging part of the plank before it begins to tip, we need to consider the rotational equilibrium of the system.

Let's analyze the forces acting on the plank from left to right:

1. The weight of the plank itself acting downward at its center, which we can call the plank's weight, Wplank.
2. The reaction force exerted by the left support, which we can call RL.
3. The reaction force exerted by the right support, which we can call RR.
4. The weight of the person walking on the overhanging part of the plank, which we can call Wperson.

Since the plank is in equilibrium (it is not rotating), the sum of the torques about any point is zero. Let's choose the right support point as our pivot for this calculation.

The torque due to the plank's weight (Wplank) about the right support is given by:

T1 = Wplank * distance from right support to the center of mass of the plank

The torque due to the reaction force RL about the right support is zero because RL acts directly through the pivot point.

The torque due to the reaction force RR about the right support is zero because RR acts directly through the pivot point.

The torque due to the person's weight (Wperson) about the right support is given by:

T2 = Wperson * distance from right support to the person

For rotational equilibrium, T1 + T2 must be equal to zero.

Now, let's plug in the given values:

The weight of the plank, Wplank, is 225 N.

The weight of the person, Wperson, is 408 N.

The length of the plank, L, is 5.0 m.

The distance from the right support to the center of mass of the plank is 0.5 * L = 0.5 * 5.0 m = 2.5 m.

The maximum distance from the right support to the person is 1.1 m (given in the problem).

Plugging in these values, the equation for rotational equilibrium becomes:

Wplank * (0.5 * L) + Wperson * 1.1 = 0

225 N * (0.5 * 5.0 m) + 408 N * 1.1 m = 0

Now, solve for the unknown variable, which is the distance x that the person can walk on the overhanging part of the plank before it begins to tip:

x = (225 N * 2.5 m) / 408 N

x ≈ 1.382 m

Therefore, a person who weighs 408 N can walk approximately 1.382 m on the overhanging part of the plank before it just begins to tip.