Forces and equilibrium — practice questions

For the particle to be in equilibrium, the vector sum of the forces must be zero.
$\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}$
$(6\mathbf{i} - 2\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) + \mathbf{F}_3 = \mathbf{0}$
$(4\mathbf{i} + 3\mathbf{j}) + \mathbf{F}_3 = \mathbf{0}$
$\mathbf{F}_3 = -4\mathbf{i} - 3\mathbf{j}$ N.\

To find the magnitude of $\mathbf{F}_3$:
$|\mathbf{F}_3| = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ N.\

To find the angle, let's first find the angle $\alpha$ the vector makes with the negative x-axis in the third quadrant.
$\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}$
$\alpha = \tan^{-1}(0.75) \approx 36.9^\circ$.
This angle is measured from the negative x-axis into the third quadrant. The angle measured anti-clockwise from the positive x-axis is $\theta = 180^\circ + 36.9^\circ = 216.9^\circ$.
Alternatively, the angle is $- (180^\circ - 36.9^\circ) = -143.1^\circ$ from the positive x-axis. Both $217^\circ$ and $-143^\circ$ (to 3 s.f.) are acceptable answers.

First, draw a free-body diagram and resolve the weight ($20g$ N) into components parallel and perpendicular to the plane.
Weight component perpendicular to plane: $W_\perp = 20g \cos(25^\circ)$.
Weight component parallel to plane: $W_\parallel = 20g \sin(25^\circ)$.\

Step 1: Find the normal reaction, R. Resolve forces perpendicular to the plane.
$\sum F_\perp = 0 \implies R - 20g \cos(25^\circ) = 0$
$R = 20(9.8) \cos(25^\circ) = 196 \times 0.9063... = 177.63...$ N.\

Step 2: Find the maximum possible frictional force, $F_{max}$.
$F_{max} = \mu R = 0.3 \times 177.63... = 53.29...$ N.\

Step 3: Find the maximum value of P ($P_{max}$). This occurs when the block is about to slide UP the plane, so friction acts DOWN the plane.
Resolve forces parallel to the plane: $\sum F_\parallel = 0$
$P_{max} - W_\parallel - F_{max} = 0$
$P_{max} = 20g \sin(25^\circ) + F_{max}$
$P_{max} = 20(9.8) \sin(25^\circ) + 53.29...$
$P_{max} = 196 \times 0.4226... + 53.29... = 82.83... + 53.29... = 136.12...$ N.\

Step 4: Find the minimum value of P ($P_{min}$). This occurs when the block is about to slide DOWN the plane, so friction acts UP the plane.
Resolve forces parallel to the plane: $\sum F_\parallel = 0$
$P_{min} + F_{max} - W_\parallel = 0$
$P_{min} = W_\parallel - F_{max}$
$P_{min} = 82.83... - 53.29... = 29.54...$ N.\

Step 5: State the final answer.
The block remains in equilibrium if $29.5 \le P \le 136$. (Values given to 3 significant figures).