A uniform plank of length 4.7 m and weight 221 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 440 N walk on the overhanging part of the plank before it just begins to tip?
To determine the maximum distance x that 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.
Here's how to approach the problem step by step:
1. Identify the forces acting on the plank:
- Weight of the plank (221 N) acting downwards at its center of mass.
- Weight of the person (440 N) acting downwards at the end of the overhanging part.
2. Determine the distances between the forces and the pivot point (fulcrum).
- For the plank's weight: The center of mass is at half of its length, so the distance from the pivot point is 4.7 m / 2 = 2.35 m.
- For the person's weight: The person is at the end of the overhanging part, so the distance from the pivot point is 1.1 m - x, where x is the distance the person can walk on the plank.
3. Apply the condition for rotational equilibrium:
The sum of the torques (moments) around the pivot point should be zero.
Mathematically, we can write: (Torque due to plank's weight) + (Torque due to person's weight) = 0
The torque τ is given by the formula: τ = force * distance
For the plank's weight: τ = 221 N * 2.35 m
For the person's weight: τ = 440 N * (1.1 m - x)
4. Set up the equation and solve for x:
221 N * 2.35 m + 440 N * (1.1 m - x) = 0
Simplify and solve for x:
518.35 N*m + 484 N*m - 440 N*x = 0
1002.35 N*m - 440 N*x = 0
440 N*x = 1002.35 N*m
x = (1002.35 N*m) / 440 N
x ≈ 2.28 m
Therefore, the person can walk approximately 2.28 meters on the overhanging part of the plank before it begins to tip.