A body of mass 10 kg is lying on a rough plane inclined at an angle of 30° to the horizontal and the coefficient of friction is 0.5. The minimum force required to pull the body up the plane is

The correct answer is 91.4 N.

When a body is pulled up a rough inclined plane, two forces resist the motion: the component of gravity along the plane (acting downward along the slope) and the kinetic friction force (also acting downward along the slope, opposing the upward motion). The applied force must overcome both of these.

Given Data:
Mass, m = 10 kg
Angle of inclination, θ = 30°
Coefficient of friction, μ = 0.5
Acceleration due to gravity, g = 9.8 m/s²

Step 1: Find the Normal Reaction Force (N)

Since there is no acceleration perpendicular to the plane, the normal force balances the perpendicular component of weight:

N=mgcos\theta

N=10\times 9.8\times cos\mathrm{30°}=98\times 0.866=84.87\text{N}

Step 2: Find the Friction Force (f)

When moving up the plane, friction acts downward along the slope:

f=\mu N=0.5\times 84.87=42.43\text{N}

Step 3: Find the Component of Weight Along the Plane

This acts downward along the slope:

mgsin\theta =10\times 9.8\times sin\mathrm{30°}=98\times 0.5=49\text{N}

Step 4: Apply Equilibrium Along the Plane

For the body to be pulled up the plane (at minimum, i.e., just about to move), the applied force F must equal the sum of all resistive forces:

F=mgsin\theta +\mu mgcos\theta

F=mg(sin\theta +\mu cos\theta )

F=10\times 9.8\times (0.5+0.5\times 0.866)

F=98\times (0.5+0.433)=98\times 0.933

F=91.43\text{N}\approx 91.4\text{N}

Conclusion: The minimum force required to pull the body up the inclined plane is 91.4 N. Note that this force is assumed to act parallel to the inclined surface. Both the gravitational pull along the slope (49 N) and the friction (42.43 N) oppose the upward motion, making the total required force approximately 91.4 N.