A body of weight 300 N is lying on a rough horizontal plane having a coefficient of friction as 0.3. Find the magnitude of the force, which can move the body, while acting at an angle of 25° with the horizontal.
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The magnitude of the force is 87.1 N
Suppose, f is the force of friction and F is the force which is applied horizontally at an angle 25°
The weight pf the body is = 300 N
The coefficient of friction (μ) = 0.3
Resolving the forces, we get,
f = F Cos 25°
⇒ f = F × 0.9063
⇒ μ R = F × 0.9063 [R is the reaction force of weight]
⇒ 0.3 × R = F × 0.9063 ....(i)
R = 300 - F Sin 25°
⇒ R = 300 - F × 0.4226 ...(ii)
By solving (i) and (ii), we get,
F = 87.1 N
The value of the force is 87.1 N.
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