Question

a person pushing a horizontal, uniformly loaded, 28.50 kg wheelbarrow of length l is attempting to get it over a step of height h = 0.410r, where r is the wheels radius. the center of gravity of the wheelbarrow is in the center of the wheelbarrow. what is the horizontal component ( p_x ) of the minimum force ( vec{p} ) necessary to push the wheelbarrow over the step? the gravitational acceleration is ( g = 9.81 m/s^2 ).

Explanation:

Step1: Find the distance from the center of the wheel to the step's corner (lever arm for gravity)

First, we find the horizontal distance \( d \) from the center of the wheel to the point where the wheel touches the step (the pivot point). Using the Pythagorean theorem in the right triangle with hypotenuse \( R \) and one leg \( R - h \):
\( d = \sqrt{R^2 - (R - h)^2} \). Substitute \( h = 0.410R \):
\( d = \sqrt{R^2 - (R - 0.410R)^2} = \sqrt{R^2 - (0.590R)^2} = R\sqrt{1 - 0.590^2} \approx R\sqrt{1 - 0.3481} = R\sqrt{0.6519} \approx 0.8074R \).

Step2: Find the lever arm for the horizontal force \( P_x \)

The vertical distance from the pivot to the center of the wheel is \( R - h = R - 0.410R = 0.590R \). For the minimum force, the torque due to \( P_x \) (about the pivot) must balance the torque due to gravity \( mg \). The torque equation is \( \tau_{P_x} = \tau_{mg} \), where \( \tau = rF\sin\theta \). For \( P_x \), the lever arm (perpendicular distance from pivot to the line of action of \( P_x \)) is \( R - h = 0.590R \), and for \( mg \), the lever arm is \( d = 0.8074R \).

Step3: Set up the torque balance equation

Torque due to \( P_x \): \( \tau_{P_x} = P_x \cdot (R - h) \) (since the force is horizontal, the perpendicular distance is vertical, \( R - h \)).
Torque due to \( mg \): \( \tau_{mg} = mg \cdot d \) (since gravity is vertical, the perpendicular distance is horizontal, \( d \)).
Setting them equal: \( P_x (R - h) = mg d \).

Step4: Solve for \( P_x \)

Substitute \( R - h = 0.590R \), \( d = 0.8074R \), and cancel \( R \) (since \( R \) appears on both sides, it cancels):
\( P_x (0.590R) = mg (0.8074R) \)
\( P_x = mg \cdot \frac{0.8074}{0.590} \).

Substitute \( m = 28.50 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \):
\( P_x = 28.50 \times 9.81 \times \frac{0.8074}{0.590} \).

Calculate \( \frac{0.8074}{0.590} \approx 1.3685 \).
Then \( 28.50 \times 9.81 \approx 279.585 \).
Multiply: \( 279.585 \times 1.3685 \approx 382 \, \text{N} \) (more precisely, \( 28.50 \times 9.81 = 279.585 \); \( 0.8074 / 0.590 \approx 1.3685 \); \( 279.585 \times 1.3685 \approx 382 \)).

Answer:

\( \boxed{382} \) (Note: More precise calculation: \( 0.8074 / 0.590 \approx 1.36847 \); \( 28.50 \times 9.81 = 279.585 \); \( 279.585 \times 1.36847 \approx 279.585 \times 1.3685 \approx 382 \, \text{N} \).)