The wheels, axle, and handles of a wheelbarrow weigh W = 67 N. The load chamber and its contents weigh WL = 511 N. The drawing shows these two forces in two different wheelbarrow designs. To support the wheelbarrow in equilibrium, the man’s hands apply a force to the handles that is directed vertically upward. Consider a rotational axis at the point where the tire contacts the ground, directed perpendicular to the plane of the paper. Find the magnitude of the man’s force for both designs.
Fh = force up of hands at x = xh from axle
Fa = force up from wheel axle at x = 0
Wb = 67 at xb from axle
Wc = 511 at xc from axle
then
Fa+Fh = 67+511
Fh * xh = 67 * xb + 511 * xc
To find the magnitude of the man's force for both designs, we need to analyze the rotational equilibrium of the wheelbarrow. In rotational equilibrium, the sum of the torques acting on the system must be zero.
Let's consider each design separately:
Design 1: The wheel is located behind the load
In this design, the weight of the wheelbarrow (W) acts downwards at the center of gravity of the wheelbarrow, and the weight of the load (WL) acts downwards at the center of gravity of the load. The man's force (F) acts upwards at the handles.
Since the rotational axis is at the point where the tire contacts the ground, the torques due to the weight of the wheel and the load are both zero because they act along the axis of rotation.
The only torque is due to the man's force (F). The distance between the handles and the rotational axis is the length of the handles (L). Therefore, the torque due to the man's force is given by the equation:
Torque = Force × Distance = F × L
For rotational equilibrium, the sum of the torques must be zero. Hence, we can write the equation:
Sum of Torques = Torque due to F - Torque due to W - Torque due to WL = 0
F × L - 0 - 0 = 0
This implies that F × L = 0.
Since the length of the handles (L) cannot be zero, the only way for F × L to be zero is if F is also zero.
Therefore, the magnitude of the man's force for Design 1 is 0.
Design 2: The wheel is located in front of the load
In this design, the weight of the wheelbarrow (W) acts downwards at the center of gravity of the wheelbarrow, and the weight of the load (WL) acts downwards at the center of gravity of the load. The man's force (F) acts upwards at the handles.
Similar to Design 1, the torques due to the weight of the wheel and the load are both zero because they act along the axis of rotation.
The only torque is due to the man's force (F). The distance between the handles and the rotational axis is the length of the handles (L). Therefore, the torque due to the man's force is given by the equation:
Torque = Force × Distance = F × L
For rotational equilibrium, the sum of the torques must be zero. Hence, we can write the equation:
Sum of Torques = Torque due to F - Torque due to W - Torque due to WL = 0
F × L - 0 - 0 = 0
This implies that F × L = 0.
Since the length of the handles (L) cannot be zero, the only way for F × L to be zero is if F is also zero.
Therefore, the magnitude of the man's force for Design 2 is also 0.
In conclusion, for both designs, the magnitude of the man's force is 0.