Answer:
Explanation:
Given
mass of crate [tex]m=50 kg[/tex]
Coefficient of static Friction [tex]\mu =0.25[/tex]
Let [tex]\theta [/tex]be angle of inclination
resolving Forces along and perpendicular to plane
Force along Plane
[tex]P\cos \theta -mg\sin \theta -f_r=0[/tex]----------1
Forces perpendicular to plane
[tex]P\sin \theta +mg\cos \theta -N=0[/tex]
[tex]f_r=\mu N[/tex]
[tex]f_r=\mu (P\sin \theta +mg\cos \theta)[/tex]
Substitute the value of [tex]f_r[/tex]
[tex]P\cos \theta -mg\sin \theta -\mu (P\sin \theta +mg\cos \theta)=0[/tex]
[tex]P\left [ \cos \theta -\mu \sin \theta \right ]=mg\left [ \sin \theta +\mu \cos \theta \right ][/tex]
[tex]P=mg\cdot \frac{\sin \theta +\mu \cos \theta }{\cos \theta -\mu sin \theta }[/tex]
(b)Minimum Force require to push the crate up the Plane
i.e. Force Must act parallel to plane
Forces Along the Plane
[tex]P-mg\sin \theta -f_r=0[/tex]-------2
Forces Perpendicular to plane
[tex]mg\cos \theta -N=0[/tex]
[tex]f_r=\mu N[/tex]
[tex]f_r=\mu mg\cos \theta [/tex]
Substitute in Equation 2
[tex]P=mg\sin \theta +f_r[/tex]
[tex]P=mg(\sin \theta +\mu \cos \theta )[/tex]
