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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.
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.
For both wheelbarrow designs, the magnitude of the man's force needed to support the wheelbarrow in equilibrium is=578NFm=578N.To find the magnitude of the man's forceFm, we need to balance the torques acting on the wheelbarrow about the point of contact between the tire and the ground. The torque due to the weight of the wheelbarrow components (wheels, axle, handles) is⋅W⋅r, whereW is the weight andr is the distance from the point of contact to the center of mass. Similarly, the torque due to the load chamber weight is⋅WL⋅r. Since the wheelbarrow is in equilibrium, the total torque must be zero.Since the torques due to the wheelbarrow components and the load chamber are acting in opposite directions, we can set up the equation⋅−⋅=0W⋅r−WL⋅r=0 to findFm. Substituting the given values forW andWL, we get67⋅−511⋅=067⋅r−511⋅r=0. Solving forr, we find=51167r=67511. Now, we substituter back into the torque equation to findFm.Thus,=+=67N+511N=578NFm=W+WL=67N+511N=578N. This force is required to support the wheelbarrow in equilibrium for both designs, ensuring that the torques due to the weight of the components and the load chamber balance each other out....