Calculate the force exerted on the handle of a wheelbarrow

[flin8812]

Homework Statement

The total weight of a wheel barrow filled with gravel=240N
Wheelbarrow held on an 18 degree incline
Other dimensions are given as in the figure provided.

Homework Equations

Assuming that the wheelbarrow is at static equilibrium,

The Attempt at a Solution

my attempt of part(a) was to have the total moment=0
where F represents the uplifting force

Mc=($$\overline{CB}$$)(F)-(rxFy)(ryFx)

= $$\sqrt{1.01^2+0.8^2}$$(0.2*240cos15)+(0.58*240sin18)

=0
but I don't know what to do next...

 

[method_man]

Try to calculate:
$$\sum x=0$$
$$\sum y=0$$
$$\sum M=0$$
I sketched some forces on your picture. I would calculate moment around C.

 

[tomcue]

I would first clarify any missing information, such as the angle of the incline and the dimensions of the wheelbarrow and handle. Assuming the incline is 18 degrees, I would then proceed to calculate the force exerted on the handle using the equation F = mg sinθ, where m is the mass of the wheelbarrow (240N) and θ is the angle of the incline (18 degrees). This would give a force of approximately 68.37N.

However, this calculation only takes into account the force exerted on the handle due to the weight of the wheelbarrow and does not consider any additional forces, such as friction or the force exerted by the person pushing the wheelbarrow. To accurately calculate the total force exerted on the handle, these factors would need to be taken into account as well. Additionally, the dimensions of the wheelbarrow and handle could affect the distribution of weight and therefore the force exerted on the handle. Further analysis and measurements may be needed to accurately determine the force exerted on the handle in this scenario.