A uniform plank of length 5.0 m and weight 231N rests horizontally on two supports, with d = 1.34m of the plank hanging over the right support. To what distance, x, can a person who weighs 477N walk on the overhanging part of the plank before it just begins to tip?
To determine the distance, x, that a person can walk on the overhanging part of the plank before it begins to tip, we need to find the point of equilibrium.
First, let's find the total weight of the plank. The weight of the plank is given as 231N.
Next, let's find the total weight acting on the overhanging part of the plank. The weight of the person is given as 477N.
Since the person is walking on the overhanging part, the weight of the person acts at a distance of x from the right support.
From the point of equilibrium, the clockwise and counterclockwise moments must balance each other.
The counterclockwise moment is given by the weight of the person multiplied by the distance x:
Counterclockwise moment = Weight of the person × distance x = 477N × x
The clockwise moment is given by the weight of the plank multiplied by the distance d:
Clockwise moment = Weight of the plank × distance d = 231N × 1.34m
For the plank to be in equilibrium, the clockwise and counterclockwise moments must be equal:
477N × x = 231N × 1.34m
To find x, rearrange the equation:
x = (231N × 1.34m) / 477N
Calculating the value of x:
x = (309.54 N·m) / 477N
x ≈ 0.648 m
Therefore, a person can walk on the overhanging part of the plank for a distance of approximately 0.648 meters before it begins to tip.
To determine the distance, x, that a person can walk on the overhanging part of the plank before it begins to tip, we need to consider the torque exerted by the person's weight and the torque exerted by the plank's weight.
First, let's calculate the torque exerted by the person's weight. The formula for torque is given by:
Torque = Force × Distance
The person's weight is given as 477N, and the distance from the right support to the point where the person is standing is x. Hence, the torque exerted by the person's weight is:
Torque_person = 477N × x
Next, let's calculate the torque exerted by the weight of the plank. The weight of the plank is given as 231N, and the distance from the right support to the center of mass (which is at the midpoint of the plank) is 5.0/2 = 2.5m. The overhanging part of the plank is d = 1.34m. Therefore, the distance from the right support to the center of mass becomes (2.5 - d).
The torque exerted by the plank's weight is:
Torque_plank = 231N × (2.5 - d)
For the plank to be in equilibrium (not tipping), the torques exerted by the person and the plank must balance each other:
Torque_person = Torque_plank
Substituting the values we have calculated:
477N × x = 231N × (2.5 - 1.34)
Now, we can solve for x:
477N × x = 231N × 1.16
x = (231N × 1.16) / 477N
x ≈ 0.563 meters
Therefore, a person can walk approximately 0.563 meters on the overhanging part of the plank before it just begins to tip.