91Ó°ÊÓ

The maximum weight that can be lifted unaided is F = 650 N.

The length of the wheelbarrow is L = 1.4 m.

The weight of the wheelbarrow is W = 80 N.

The distance of the center of gravity of the wheelbarrow from the center of the wheel is a = 0.50 m.

The distance of the center of gravity of load from the center of the wheel is a = 0.50 m.

When a body having a certain weight rests on the ground, then a reaction force equal to the body’s weight acts on it, and this force is described as ‘normal reaction force’.

The direction of the reaction force is always normal to the ground.

$W_1$The free-body diagram of the wheelbarrow is given by,

Here, N is the normal reaction on the wheel of the wheelbarrow and is the load carried by the wheelbarrow.

Taking moment about the right end of the wheelbarrow, and we get,

$F×L-\left(W+W_1\right)×a=0$

Substituting the values in the above expression and we get:

$$\begin{gathered} 650\text{N}×\text{1.4}\text{m}-\left(80\text{N}+W_1\right)×0.50\text{m}=0 \\ 910\text{N}·\text{m}-40\text{N}·\text{m}-0.50W_1\text{m}=0 \end{gathered}$$

Solving further as:

$$\begin{gathered} 0.50W_1\text{m}=870\text{N}·\text{m} \\ {W}_1=\frac{870\text{N}·\text{m}}{0.50\text{m}} \\ {W}_1=1740\text{N} \end{gathered}$$

Hence, the load lifted using the wheelbarrow is 1740 N.

By using the wheelbarrow, the value of the load lifted is 1740 N which is greater than the load lifted unaided.

The normal reaction on the wheel of the wheelbarrow acting upwards is given by:

$N=\left(W+W_1\right)-F$

Substituting the values in the above expression and we get:

$$\begin{gathered} N=\left(80\text{N}+1740\text{N}\right)-650\text{N} \\ N=1820\text{N}-650\text{N} \\ N=1170\text{N} \end{gathered}$$

Here, the force that enables to lift more than 650 N using the wheelbarrow is because of the normal reaction acting on the wheel of the wheelbarrow in the upward direction.

Hence, the force that enables to lift more than 650 N is the normal reaction acting on the wheel of the wheelbarrow.