(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 ([tex]F[/tex]), in newtons:
[tex]F = \frac{\mu_{s}\cdot m\cdot g}{\cos \theta} [/tex] (1)
Where:
- [tex]\mu_{s}[/tex] - Coefficient of static friction, no unit.
- [tex]m[/tex] - Mass of the particle, in kilograms.
- [tex]g[/tex] - Gravitational acceleration, in meters per square second.
- [tex]\theta [/tex] - Angle of the external force with respect to the plane, in degrees.
(i) If we know that [tex]\mu_{s} = 0.5[/tex] [tex]m = 0.8\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}} [/tex] and [tex]\theta = 0^{\circ}[/tex], then the pull force is:
[tex]F = \frac{(0.5)\cdot (0.8\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)}{\cos 0^{\circ}}[/tex]
[tex]F = 4\,N[/tex]
The least pull force if the force is horizontal is 4 newtons. [tex]\blacksquare[/tex]
(ii) If we know that [tex]\mu_{s} = 0.5[/tex] [tex]m = 0.8\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}} [/tex] and [tex]\theta = 30^{\circ}[/tex], then the pull force is:
[tex]F = \frac{(0.5)\cdot (0.8\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)}{\cos 30^{\circ}}[/tex]
[tex]F = 4.619\,N[/tex]
The least pull force if the force is at an angle of 30° to the plane is 4.619 newtons. [tex]\blacksquare[/tex]
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