Answer :
Answer:
The horizontal component of the minimum force is 144.24 N.
Explanation:
Given that,
Loaded = 23.90 kg
Height = 0.370 R
Where, R=wheel's radius
We need to calculate the acceleration
Using formula of acceleration
[tex]R^2=a^2+(R-h)^2[/tex]
[tex]a^2=R^2-(R-h)^2[/tex]
Put the value into the formula
[tex]a^2=R^2-(R-0.370R)^2[/tex]
[tex]a=\sqrt{0.631R^2}[/tex]
[tex]a=0.776R[/tex]
We need to calculate the horizontal component of the minimum force
Using moment about center of point of contact
[tex]P_{x}(R-h)=\dfrac{mg}{2}\times a[/tex]
[tex]P_{x}(R-0.370)=\dfrac{23.90\times9.8}{2}\times0.776R[/tex]
[tex]P_{x}=\dfrac{23.90\times9.8\times0.776R}{2(R-0.370)}[/tex]
[tex]P_{x}=144.24\ N[/tex]
Hence, The horizontal component of the minimum force is 144.24 N.