Question

A person is pushing a wheelbarrow along a ramp that makes an angle α = 39.0° with the horizontal. The wheelbarrow and load have a combined mass of 29.35 kg with the center of mass at the midpoint of the length L. What is the total force the person must apply so that the wheelbarrow remains horizontal on that ramp?

Answer 1

`F=185.006N`

Step-by-step explanation:

Summing moments about the wheelbarrow-to-ramp contact point to zero

∑
`F=F_y*L-F_N`

`F_y[L]-29.35kg*(9.81)*[(L)/(2)]=0`

`Fy=(m*g)/(2)=(29.35kg*9.8m/s^2)/(2)=143.8N`

Summing forces in the vertical direction to zero

`F_N*cos(36) + Fy - 29.35kg(9.8m/s^2)=0`

`F_N*cos(39)= 29.35kg(9.8m/s^2)-143.8N`

`F_N=185.07N`

Summing horizontal forces to zero

`F_N*sin(39)- F_x=0`

`F_x=116.47N`

total force

`F=√(F_x^2+F_y^2)=√(116.47^2+143.8^2)`

`F=185.006N`

which is interestingly at an angle of 36.0° from the vertical.