Question

In this question take g to be 10 ms^-2
A particle of mass 0.8 kg is at rest on a rough horizontal plane. The coefficient of friction between the particle and the plane is 0.5. Find the least force required to pull the particle along the plane if the force is
(i) horizontal
(ii) at an angle of 30° to the plane.

Answer 1

(i) The least pull force if the force is horizontal is 4 newtons.

(ii) The least pull force if the force is at an angle of 30° to the plane is 4.619 newtons.

Determination of the least force required to pull a mass

By Newton's second law and definition of the maximum kinetic friction we derive an equation for the least force required to pull the particle along the plane ($F$), in newtons:

$F = \frac{\mu_{s}\cdot m\cdot g}{\cos \theta}$ (1)

Where:

• $\mu_{s}$ - Coefficient of static friction, no unit.
• $m$ - Mass of the particle, in kilograms.
• $g$ - Gravitational acceleration, in meters per square second.
• $\theta$ - Angle of the external force with respect to the plane, in degrees.

(i) If we know that $\mu_{s} = 0.5$ $m = 0.8\,kg$, $g = 10\,\frac{m}{s^{2}}$ and $\theta = 0^{\circ}$, then the pull force is:

$F = \frac{(0.5)\cdot (0.8\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)}{\cos 0^{\circ}}$

$F = 4\,N$

The least pull force if the force is horizontal is 4 newtons. $\blacksquare$

(ii) If we know that $\mu_{s} = 0.5$ $m = 0.8\,kg$, $g = 10\,\frac{m}{s^{2}}$ and $\theta = 30^{\circ}$, then the pull force is:

$F = \frac{(0.5)\cdot (0.8\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)}{\cos 30^{\circ}}$

$F = 4.619\,N$

The least pull force if the force is at an angle of 30° to the plane is 4.619 newtons. $\blacksquare$

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