A uniform plank of length 5.1 m and weight 235 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 458 N walk on the overhanging part of the plank before it just begins to tip?
this is the Drawing:
0 (person)
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^ ^ (d = 1.1 from lever to edge)
Thank you for trying to explain the figure. I think I get the picture.
One can walk the plank until the force on the leftmost support is zero.
Set the moment about the rightmost pivot equal to zero and assume zero force at that location. You should be able to solve for the distance x where the person stands. Remember to include the moment due to the weight of the plank, acting at the center of mass.
To solve this problem, we need to understand the concept of torques and equilibrium. Torque is the rotational equivalent of force and is calculated by multiplying the applied force by the distance from the fulcrum.
In this case, the fulcrum is the left support, and the weight of the plank and the person (combined weight) act downward through the center of gravity, which is the midpoint of the plank.
Let's calculate the torque due to the weight of the plank. The distance of the plank's center of gravity from the left support is half of its total length, which is 5.1 m / 2 = 2.55 m. The weight of the plank is given as 235 N. Therefore, the torque due to the weight of the plank is 235 N * 2.55 m = 599.25 Nm.
Next, let's calculate the torque due to the person. The person's weight is given as 458 N, and their position from the left support is the total length of the plank minus the overhang distance, which is 5.1 m - 1.1 m = 4.0 m. Therefore, the torque due to the person is 458 N * 4.0 m = 1832 Nm.
For the plank to be in equilibrium, the sum of all torques acting on it should be zero. Since the torques due to the weight of the plank and the person act in opposite directions, we can write the equation:
599.25 Nm + (-1832 Nm) + Σtorque_other = 0
Simplifying the equation, we have:
Σtorque_other = 1832 Nm - 599.25 Nm = 1232.75 Nm
To prevent the plank from tipping, the torque due to any other external forces acting on it must be less than or equal to 1232.75 Nm. In this case, the only other external force is the weight of the person walking on the overhanging part of the plank.
Let's assume the person walks a distance x on the overhang. The distance from the left support to the person's location is 1.1 m + x. Therefore, for equilibrium, the torque due to the person's weight acting on the right side of the support should also be less than or equal to 1232.75 Nm.
The torque due to the person's weight on the right side is 458 N * (1.1 m + x). So we can set up the equation:
458 N * (1.1 m + x) ≤ 1232.75 Nm
Now we solve for x:
458 N * 1.1 m + 458 N * x ≤ 1232.75 Nm
503.8 N + 458 N * x ≤ 1232.75 Nm
458 N * x ≤ 1232.75 Nm - 503.8 N
458 N * x ≤ 728.95 Nm
x ≤ 728.95 Nm / 458 N
x ≤ 1.59 m
Therefore, a person weighing 458 N can walk on the overhanging part of the plank, up to a distance of 1.59 m before it just begins to tip.