A body of mass 6kg rests on an inclined plane. The normal reaction is R, and the limiting frictional force is F as shown in the image above. if F = 30N, and g = 10ms\(^{-2}\) then the angle of inclination \( \theta \) is
15\(^o\)
30\(^o\)
45\(^o\)
60\(^o\)
75\(^o\)
Explanation
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SINTITAN = F/MG
= 30/10 x 6
= 30/60
= 0.5
THEREFORE, SIN INVERSE OF 0.5 WILL BE 30 DEGREES
it's always confusing
thought that if the angle is to the horizontal it's cos
and to the vertical is sin
To find the angle of inclination θ, we’ll resolve forces acting on the body resting on the inclined plane.
Given:
• Mass of body, m = 6 kg
• Gravitational acceleration, g = 10 m/s²
• Weight, W = mg = 6 × 10 = 60 N
• Limiting frictional force, F = 30 N
Step 1: Understand the forces
On an inclined plane:
• Weight (W) acts vertically downward.
• The component of weight perpendicular to the plane:
W⊥ = W cosθ = 60 cosθ
• The component of weight parallel to the plane (which causes sliding):
W‖ = W sinθ = 60 sinθ
• Normal reaction R = W cosθ
• Frictional force F = μR, and at limiting condition, F = 30 N
But at limiting equilibrium:

So,


Step 2: Find θ

Final Answer: B. 30°
