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 mathrm{~m} / mathrm{s}^2 ).

Explanation:

Step1: Find the perpendicular distance from the pivot (edge of the step) to the line of action of the weight force

The center - of - gravity of the wheelbarrow is at its center. The distance from the pivot (edge of the step) to the center of the wheel (where the weight acts effectively, assuming uniform loading) in the vertical direction for torque calculation:
Let's first find the distance \(d_{mg}\) from the pivot to the line of action of the weight force. Using the Pythagorean theorem, if the radius is \(R\) and \(h = 0.410R\), then the vertical distance from the center of the wheel (where the weight acts) to the pivot is \(y=\sqrt{R^{2}-(R - h)^{2}}\). Substitute \(h = 0.410R\) into it:
\[ \begin{align*} y&=\sqrt{R^{2}-(R - 0.410R)^{2}}\\ &=\sqrt{R^{2}-(0.59R)^{2}}\\ &=\sqrt{R^{2}-0.3481R^{2}}\\ &=\sqrt{0.6519R^{2}}\\ &= 0.807R \end{align*} \]
The weight of the wheelbarrow \(W=mg\), where \(m = 28.50\space kg\) and \(g = 9.81\space m/s^{2}\), so \(W=28.50\times9.81=279.585\space N\)

Step2: Find the perpendicular distance from the pivot to the line of action of the applied force \(P\)

The distance \(d_{P}\) from the pivot to the line of action of the force \(P\) (assuming the force is applied at the end of the wheelbarrow of length \(L\)). When the wheelbarrow is just about to lift over the step, the torque due to the weight \(W\) about the pivot is balanced by the torque due to the force \(P\).
The torque \(\tau= rF\sin\theta\). For the weight, \(\tau_{mg}=W\times d_{mg}\), and for the force \(P\), assume the force is applied horizontally. The perpendicular distance from the pivot to the line of action of \(P\) (if the length of the wheelbarrow is \(L\)): Let's assume the wheelbarrow is just about to pivot. The vertical distance from the pivot to the line of action of \(P\) (using similar geometric relations as above for the maximum - effort case) is \(d_{P}=L\) (assuming the length \(L\) is measured from the end where the force is applied to the pivot - like point at the edge of the step in the torque - balance scenario)

Step3: Set up the torque - balance equation

At the minimum force condition (when the wheelbarrow is just about to lift over the step), the torque due to the weight about the pivot is equal to the torque due to the applied force about the pivot. \(\tau_{mg}=\tau_{P}\)
\(W\times d_{mg}=P\times d_{P}\)
We want to find \(P_{x}\) (since the force \(P\) is horizontal, \(P = P_{x}\)). Rearranging for \(P_{x}\): \(P_{x}=\frac{W\times d_{mg}}{d_{P}}\)
Assume \(d_{mg} = 0.807R\) and if we assume the length of the wheel - barrow \(L\) (for the torque - balance, the ratio of distances is what matters. If we assume the geometric ratio of the torque - arms, and since we are not given \(L\) in terms of \(R\) explicitly, but using the fact that in the torque - balance \(\frac{P_{x}}{W}=\frac{d_{mg}}{d_{P}}\). If we assume the wheel - barrow length \(L\) (from the end where the force is applied to the pivot - point at the step) and using the geometric relations for the torque - arms.
Another approach: Using the fact that for the wheel (a circular shape), when calculating the torque about the step - edge (pivot):
The torque due to the weight \(W\) about the step - edge (pivot) is \(W\times\sqrt{R^{2}-(R - h)^{2}}\) and the torque due to \(P_{x}\) is \(P_{x}\times\sqrt{(L)^{2}+(R - h)^{2}}\) (assuming \(L\) is the horizontal distance from the point of force application to the vertical line through the pivot). But if we consider the wheel - only (assuming the wheel - barrow length \(L\) is such that the torque - balance is based on the wheel's geometry for the critical case.
Using the torque - balance \(\sum\tau = 0\), \(P_{x}\times\sqrt{R^{2}-(R - h)^{2}}=W\times(R - h)\)
Substitute \(h = 0.410R\)
\[ \begin{align*} P_{x}&=W\times\frac{R - h}{\sqrt{R^{2}-(R - h)^{2}}}\\ &=279.585\times\frac{R-0.410R}{\sqrt{R^{2}-(R - 0.410R)^{2}}}\\ &=279.585\times\frac{0.59R}{\sqrt{R^{2}-0.3481R^{2}}}\\ &=279.585\times\frac{0.59}{0.807}\\ &=204\space N \end{align*} \]

Answer:

\(204\space N\)