Answer:
The minimum force is 357.12 N.
Explanation:
Given that,
Load = 25.30 kg
Height h = 0.430 R
We need to calculate the height
[tex]h=(R-0.430 R)[/tex]
[tex]h=(1-0.430)R[/tex]
[tex]h=0.57R[/tex]
We need to calculate the value of x
Using formula of x
[tex]x=\sqrt{R^2-(0.57R)}[/tex]
[tex]x=0.821R[/tex]
We need to calculate the horizontal force
Using balance equation of torque
[tex]mgx=F\times h[/tex]
[tex]F=\dfrac{mgx}{h}[/tex]
Put the value into the formula
[tex]F=\dfrac{25.30\times9.8\times0.821 R}{0.57 R}[/tex]
[tex]F=357.12\ N[/tex]
Hence, The minimum force is 357.12 N.
The horizontal component of the minimum force required to push the wheel barrow over the given step is zero.
The given parameters:
The minimum force required to push the wheel barrow over the given step is calculated as follows;
F = mg
The vertical component of the minimum force is calculated as follows;
Fy = mg sin(θ)
The horizontal component of the minimum force required to push the wheel barrow over the given step is calculated as follows;
Fx = mg cos(θ)
where;
θ is the angle between the radius of the wheelbarrow and the vertical height = 90⁰.
Thus, the horizontal component of the minimum force required to push the wheel barrow over the given step is zero.
Learn more about horizontal forces here: https://brainly.com/question/12979825
