A boy weighing 650 N wants to cross a creek by using a 5.00 meter long plank as a bridge. He places the plank across the creek and begins to walk across. Consider the supporting forces to be at the ends of the plank, and the plank itself to be weightless. At one point, the supporting force at the end where the boy is going is 250 N. Where is the boy located when this is the case?
sum moments about the boy, located at x from the end where he is going.
250*x=(650-250)*(5-x)
solve for x
To determine the location of the boy when the supporting force at one end of the plank is 250 N, we can use the principle of torque equilibrium.
Torque is a rotational force that is applied at a distance from the point of rotation. In this case, we can consider the point at which the plank is resting on the ground as the rotation point.
The torque exerted by the boy can be calculated using the formula:
Torque = Force × Distance
Since the boy's weight is acting downwards, we need to balance it with the supporting force at the other end of the plank. At equilibrium, the torques exerted by the boy and the supporting force should be equal.
Let's assume that the boy's position is denoted by 'x' meters from the end of the plank where the supporting force is 250 N. The distance from this end to the rotation point (fulcrum) is (5 - x) meters.
The torque exerted by the boy can be calculated as follows:
Torque exerted by the boy = Weight of the boy × Distance between the boy and the rotation point
= 650 N × (5 - x) meters
Now, we can set this torque equal to the torque exerted by the supporting force:
650 N × (5 - x) = 250 N × x
Simplifying the equation:
3250 - 650x = 250x
Combining like terms:
3250 = 900x
Dividing both sides by 900:
x = 3250 / 900
x ≈ 3.61 meters
Therefore, the boy is located approximately 3.61 meters from the end of the plank where the supporting force is 250 N.