Question

A person is pushing a fully loaded 24.50 kg wheelbarrow at constant velocity along a plank that makes an angle =42.0∘ with the horizontal. The load is distributed such that the center of mass of the wheelbarrow is exactly halfway along its length . What is the magnitude of the total force the person must apply so that the wheelbarrow is parallel to that plank? You may neglect the radius of the wheel in your analysis. The gravitational acceleration is =9.81 m/s2.

Answer 1

Answer:

184 N

Explanation:

Draw a free body diagram of the wheelbarrow.  There are three forces on the wheelbarrow:

Weight force mg pulling straight down, at the center.

Normal force N pushing perpendicular to the plank, at the wheel.

Applied force F at the handle.

Take x direction to be parallel to the plank, and the y direction to be perpendicular to the plank.

Sum of torques about the wheel:

∑τ = Iα

Fᵧ L − (mg cos α) (L/2) = 0

Fᵧ L = (mg cos α) (L/2)

Fᵧ = mg cos α / 2

Fᵧ = (24.50 kg) (9.81 m/s²) (cos 42.0°) / 2

Fᵧ = 89.3 N

Sum of forces in the x direction:

∑F = ma

Fₓ − mg sin α = 0

Fₓ = mg sin α

Fₓ = (24.50 kg) (9.81 m/s²) (sin 42.0°)

Fₓ = 160.8 N

So the magnitude of the force is:

F = √(Fₓ² + Fᵧ²)

F = √((160.8 N)² + (89.3 N)²)

F = 184 N

Answer 2

We have that for the Question, it can be said that

From the question we are told

A person is pushing a fully loaded 24.50 kg wheelbarrow at constant velocity along a plank that makes an angle =42.0∘ with the horizontal. The load is distributed such that the center of mass of the wheelbarrow is exactly halfway along its length . What is the magnitude of the total force the person must apply so that the wheelbarrow is parallel to that plank? You may neglect the radius of the wheel in your analysis. The gravitational acceleration is =9.81 m/s2.

Generally the equation for the Force  is mathematically given as

For x axis

$F_x=mgsin\alpha\\\\F_x=24.5 *sin42 *9.8\\\\F_x=160.65N\\\Fxl-(mgcos\theta)1/2=0$

$F_y=\frac{mgcos\alpha}{2}\\\\F_y=\frac{24.5*9.8*cos42}{2}\\\\F_y=89.2N$

$F_net=\sqrt{f_x^2+F_y^2}\\\\F_net=\sqrt{160.65^2+89.2^2}$

F_net=183.75

Therefore

the magnitude of the total force the person must apply so that the wheelbarrow is parallel to that plank is

F_net=183.75

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