Past Exam Questions

(i) Using the cosine rule in the triangle formed by the forces:

\(X = 14 - 13 \cos \theta\)

\(Y = 13 \sin \theta\)

\(14^2 + 13^2 - 2 \times 13 \times 14 \cos \theta = 15^2\)

Solving for \(\theta\), we find \(\theta = 67.4^\circ\).

(ii) The component of the resultant in the direction of the 14 N force is:

\(X = 15 \cos \left( \arctan \left( \frac{Y}{X} \right) \right)\)

The component is 9 N.

First, resolve the forces into their components along the i and j directions.

For the 31 N force, it acts entirely in the i direction: \(X_1 = 31\).

For the 26 N force, resolve using \(\tan \alpha = \frac{5}{12}\):

\(\cos \alpha = \frac{12}{13}\) and \(\sin \alpha = \frac{5}{13}\).

Thus, \(X_2 = 26 \cos \alpha = 26 \times \frac{12}{13} = 24\).

\(Y_2 = 26 \sin \alpha = 26 \times \frac{5}{13} = 10\).

For the 58 N force, it acts entirely in the j direction: \(Y_1 = 58\).

Sum the components:

\(X = 31 + 24 = 55\).

\(Y = 58 - 10 = 48\).

Calculate the magnitude of the resultant force:

\(R = \sqrt{X^2 + Y^2} = \sqrt{55^2 + 48^2} = 73 \text{ N}\).

Calculate the direction of the resultant force:

\(\tan \theta = \frac{Y}{X} = \frac{48}{55}\).

\(\theta = \arctan \left( \frac{48}{55} \right) \approx 41.1^\circ\).

Thus, the resultant is 73 N and the direction is 41.1° to the i direction.

(i) To find the component of the resultant in the direction of AB:

The forces are resolved along AB using the cosine of the angles given:

\(2 \times 12 \cos 40^\circ - 15 \cos 50^\circ = 8.74 \text{ N}\)

(ii) To find the component perpendicular to AB:

The forces are resolved perpendicular to AB using the sine of the angles:

\(12 \sin 40^\circ + 12 \sin 40^\circ - 15 \sin 50^\circ = 11.5 \text{ N}\)

(ii) To find the magnitude and direction of the resultant:

The magnitude of the resultant force is given by:

\(R = \sqrt{(8.74)^2 + (11.5)^2} = 14.4 \text{ N}\)

The direction \(\theta\) is given by:

\(\tan \theta = \frac{11.5}{8.74}\)

Thus, \(\theta = 52.7^\circ\) anticlockwise from the i direction.

To solve for \(F\) and \(\theta\), resolve the forces into horizontal and vertical components:

Horizontal: \(F \cos \theta = 12 \cos 30^\circ\)

Vertical: \(F \sin \theta = 10 - 12 \sin 30^\circ\)

Calculate \(F\) and \(\theta\):

\(F \cos \theta = 10.932\)

\(F \sin \theta = 4\)

Using \(F^2 = X^2 + Y^2\):

\(F = \sqrt{10.932^2 + 4^2} = 11.1 \text{ N}\)

Using \(\tan \theta = \frac{Y}{X}\):

\(\theta = \arctan \left( \frac{4}{10.932} \right) = 21.1^\circ\)

For part (ii), when the 12 N force is removed, the resultant force is the vector sum of the remaining forces:

Magnitude is 12 N, direction is 30° clockwise from the positive x-axis.

(i) To find \(F\), use the Pythagorean theorem for the components:

\(F^2 = 27.5^2 + (-24)^2\)

\(F = \sqrt{27.5^2 + 24^2} = 36.5\)

To find \(\alpha\), use the tangent function:

\(\tan \alpha = \frac{-24}{27.5}\)

\(\alpha = \arctan\left(\frac{-24}{27.5}\right) = 41.1^\circ\)

(ii) The second force acts at 90° anticlockwise, so the angle between the forces is \(\alpha + 90^\circ\). The magnitude of the resultant force \(R\) is given by:

\(R = \sqrt{F^2 + 87.6^2 + 2 \cdot F \cdot 87.6 \cdot \cos(90^\circ + \alpha)}\)

\(R = \sqrt{36.5^2 + 87.6^2 + 2 \cdot 36.5 \cdot 87.6 \cdot \cos(90^\circ + 41.1^\circ)} = 94.9\)

To find \(\theta\), use the tangent function:

\(\tan \theta = \frac{87.6 \cdot \sin(48.9^\circ) - 24}{27.5 + 87.6 \cdot \cos(48.9^\circ)}\)

\(\theta = \arctan\left(\frac{87.6 \cdot \sin(48.9^\circ) - 24}{27.5 + 87.6 \cdot \cos(48.9^\circ)}\right) = 26.3^\circ\)

To find \(\alpha\), we use the vertical component equation:

\(7.3 \sin \alpha = 5.5\)

Solving for \(\alpha\):

\(\sin \alpha = \frac{5.5}{7.3}\)

\(\alpha = \sin^{-1}\left(\frac{5.5}{7.3}\right) \approx 48.9^\circ\)

To find the magnitude of the resultant, we use the horizontal component equation:

\(R = 6.8 - 7.3 \cos 48.9^\circ\)

Calculating \(R\):

\(R \approx 6.8 - 7.3 \times 0.662\)

\(R \approx 6.8 - 4.835\)

\(R \approx 2 \text{ N}\)

First, resolve the forces in the x-direction and y-direction:

\(X = 160 + 250 \cos \alpha = 160 + 250 \times 0.96 = 400 \text{ N}\)

\(Y = 370 - 250 \sin \alpha = 370 - 250 \times 0.28 = 300 \text{ N}\)

Calculate the magnitude of the resultant force \(R\) using Pythagoras' theorem:

\(R = \sqrt{X^2 + Y^2} = \sqrt{400^2 + 300^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \text{ N}\)

Calculate the angle \(\theta\) that the resultant makes with the x-direction using the tangent function:

\(\tan \theta = \frac{Y}{X} = \frac{300}{400} = 0.75\)

\(\theta = \arctan(0.75) \approx 36.9^\circ\)

To find the value of \(Q\), we resolve the forces and use trigonometric identities. The angles between the forces and the resultant are given as 40° and 80°.

Using the cosine rule for the triangle formed by the forces and the resultant:

\(Q - P \cos 60^{\circ} = 12 \cos 80^{\circ}\)

\(P \sin 60^{\circ} = 12 \sin 80^{\circ}\)

From the first equation, we can express \(Q\) in terms of \(P\):

\(Q \cos 80^{\circ} + P \cos 40^{\circ} = 12\)

\(P \sin 40^{\circ} = Q \sin 80^{\circ}\)

Eliminating \(P\) from these equations gives:

\(Q \cos 80^{\circ} + \frac{12 \sin 80^{\circ} \cos 60^{\circ}}{\sin 60^{\circ}} = 12 \cos 80^{\circ}\)

Solving for \(Q\):

\(Q = 8.91\)

(a) To find the values of \(F\) and \(G\) when the forces are in equilibrium, resolve the forces vertically and horizontally.

Vertically: \(G \sin 60^\circ + 2F \sin 40^\circ - 10 = 0\)

Horizontally: \(F + G \cos 60^\circ - 2F \cos 40^\circ = 0\)

Solving these equations simultaneously gives \(F = 4.53\) and \(G = 4.82\).

(b) Given \(F = 3\), resolve the forces parallel to the 10N force and equate this expression to zero:

\(G \sin 60^\circ + 2 \times 3 \sin 40^\circ - 10 = 0\)

Solving for \(G\) gives \(G = 7.09\) to 3 significant figures.

To find the components of the resultant force, we resolve each force into its x and y components.

For the 7 N force along the x-axis:

\(F_{x1} = 7 \text{ N}\)

For the 10 N force at 50° to the x-axis:

\(F_{x2} = 10 \cos 50^{\circ}\)

\(F_{y2} = 10 \sin 50^{\circ}\)

For the 15 N force at 80° to the x-axis:

\(F_{x3} = -15 \cos 80^{\circ}\)

\(F_{y3} = 15 \sin 80^{\circ}\)

The negative sign for \(F_{x3}\) is because the force is acting in the negative x-direction.

Summing the x-components:

\(X = F_{x1} + F_{x2} + F_{x3} = 7 + 10 \cos 50^{\circ} - 15 \cos 80^{\circ}\)

Summing the y-components:

\(Y = F_{y2} + F_{y3} = 10 \sin 50^{\circ} + 15 \sin 80^{\circ}\)

Calculating these gives:

\(X = 10.8 \text{ N}\)

\(Y = 22.4 \text{ N}\)

To find the direction of the resultant force, we use:

\(\theta = \arctan \left( \frac{Y}{X} \right) = \arctan \left( \frac{22.4}{10.8} \right)\)

Calculating this gives:

\(\theta = 64.2^{\circ}\)

The direction is 64.2° anticlockwise from the x-axis.

(i) To find the components of the 8 N force:

1. Parallel to the 10 N force: The component is \(8 \cos \theta\). Therefore, the resultant component parallel to the 10 N force is \(10 - 8 \cos \theta\).
2. Perpendicular to the 10 N force: The component is \(8 \sin \theta\).

(ii) The magnitude of the resultant force is given as 8 N. Using the Pythagorean theorem for the components:

\((10 - 8 \cos \theta)^2 + (8 \sin \theta)^2 = 8^2\)

Expanding and simplifying:

\(100 - 160 \cos \theta + 64 \cos^2 \theta + 64 \sin^2 \theta = 64\)

Using \(\sin^2 \theta + \cos^2 \theta = 1\):

\(100 - 160 \cos \theta + 64 = 64\)

\(164 - 160 \cos \theta = 64\)

\(160 \cos \theta = 100\)

\(\cos \theta = \frac{5}{8}\)

(i) Since the particle is in equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

Resolve the forces horizontally and vertically:

Horizontal: \(7 = F \cos \theta\)

Vertical: \(4 = F \sin \theta\)

Using \(F^2 = 7^2 + 4^2\), we find:

\(F = \sqrt{49 + 16} = \sqrt{65} \approx 8.06\)

Using \(\tan \theta = \frac{4}{7}\), we find:

\(\theta = \arctan\left(\frac{4}{7}\right) \approx 29.7^\circ\)

(ii) When the 7 N force is removed, the resultant force is the vector sum of the remaining forces:

The magnitude is 7 N, and the direction is opposite to the original 7 N force.

(i) To find \(\theta\), use the component of the resultant in the direction OA:

\(8 + 8 \cos \theta = 9\)

Solving for \(\cos \theta\):

\(8 \cos \theta = 1\)

\(\cos \theta = \frac{1}{8}\)

\(\theta = \cos^{-1} \left( \frac{1}{8} \right) \approx 82.8^\circ\)

(ii) To find the magnitude of the resultant, use the cosine rule in the triangle formed by the forces and the resultant:

\(R^2 = 8^2 + 8^2 + 2 \times 8 \times 8 \cos \theta\)

\(R^2 = 64 + 64 + 128 \cos \theta\)

Using \(\cos \theta = \frac{1}{8}\):

\(R^2 = 128 + 128 \times \frac{1}{8}\)

\(R^2 = 128 + 16\)

\(R^2 = 144\)

\(R = \sqrt{144} = 12 \text{ N}\)

(i) To find P, use the formula for the magnitude of a vector:

\(P = \sqrt{(-2.8)^2 + 9.6^2} = \sqrt{7.84 + 92.16} = \sqrt{100} = 10 \text{ N}\).

To find R, use the Pythagorean theorem for the resultant of perpendicular forces:

\(R = \sqrt{10^2 + 25^2} = \sqrt{100 + 625} = \sqrt{725} \approx 26.9 \text{ N}\).

(ii) To find α, use the tangent function:

\(\tan \alpha = \frac{9.6}{2.8}\), so \(\alpha = \arctan\left(\frac{9.6}{2.8}\right) \approx 73.7^\circ\).

For the components of the 25N force:

(a) x-direction: \(25 \cos(90^\circ - \alpha) = 25 \sin(\alpha) \approx 24 \text{ N}\).

(b) y-direction: \(25 \sin(90^\circ - \alpha) = 25 \cos(\alpha) \approx 7 \text{ N}\).

(iii) To find θ, use the tangent function for the resultant vector:

\(\tan \theta = \frac{7 + 9.6}{24 - 2.8}\), so \(\theta = \arctan\left(\frac{16.6}{21.2}\right) \approx 38.1^\circ\).

To find the resultant of the three forces, we resolve each force into its horizontal ( X ) and vertical ( Y ) components.

For the 7 N force (acting horizontally):

\(X_1 = 7, \quad Y_1 = 0\)

For the 5 N force (at 50° to the horizontal):

\(X_2 = 5 \cos 50°, \quad Y_2 = 5 \sin 50°\)

For the 6 N force (at 30° below the horizontal):

\(X_3 = -6 \cos 30°, \quad Y_3 = -6 \sin 30°\)

Summing the components:

\(X = X_1 + X_2 + X_3 = 7 + 5 \cos 50° - 6 \cos 30°\)

\(Y = Y_1 + Y_2 + Y_3 = 5 \sin 50° - 6 \sin 30°\)

Calculating these values:

X \approx 5.02, \quad Y \approx 0.83

The magnitude of the resultant force is given by:

\(R = \sqrt{X^2 + Y^2} \approx \sqrt{5.02^2 + 0.83^2} \approx 5.09 \text{ N}\)

The direction of the resultant force relative to the 7 N force is given by:

\(\theta = \tan^{-1} \left(\frac{Y}{X}\right) \approx \tan^{-1} \left(\frac{0.83}{5.02}\right) \approx 9.4°\)

Thus, the magnitude of the resultant force is 5.09 N, and its direction is 9.4° anti-clockwise from the force of magnitude 7 N.

To find the resultant force \(R\) and the angle \(\alpha\), we resolve the forces into their horizontal and vertical components.

Horizontal component, \(X = 100 + 250 \cos 70^\circ\)

Vertical component, \(Y = 300 - 250 \sin 70^\circ\)

Calculate \(X\) and \(Y\):

\(X = 100 + 250 \times 0.3420 = 185.5\)

\(Y = 300 - 250 \times 0.9397 = 65.1\)

The magnitude of the resultant force \(R\) is given by:

\(R^2 = X^2 + Y^2\)

\(R^2 = 185.5^2 + 65.1^2\)

\(R = \sqrt{185.5^2 + 65.1^2} = 197\)

The angle \(\alpha\) is given by:

\(\tan \alpha = \frac{Y}{X} = \frac{65.1}{185.5}\)

\(\alpha = \arctan \left( \frac{65.1}{185.5} \right) = 19.3^\circ\)

To solve this problem, we resolve the forces into components along and perpendicular to the direction of PQ.

(i) (a) For resolving in the direction PQ:

The component is given by:

\(2 \times 10 \cos 30^\circ - 6 \cos 60^\circ\)

\(= 20 \times \frac{\sqrt{3}}{2} - 6 \times \frac{1}{2}\)

\(= 10\sqrt{3} - 3\)

\(\approx 14.3 \text{ N}\)

(i) (b) For resolving perpendicular to PQ:

The component is given by:

\(\pm 6 \cos 30^\circ - 6 \cos 60^\circ\)

\(= \pm 6 \times \frac{\sqrt{3}}{2} - 6 \times \frac{1}{2}\)

\(= \pm 3\sqrt{3} - 3\)

\(\approx \pm 5.20 \text{ N}\)

(ii) To find the magnitude of the resultant of the three forces:

Use the formula:

\(\sqrt{(\text{component in direction of } PQ)^2 + (\text{component perpendicular to } PQ)^2}\)

\(= \sqrt{(14.3)^2 + (5.20)^2}\)

\(\approx 15.2 \text{ N}\)

First, resolve the forces horizontally and vertically. Let the horizontal component be \(X\) and the vertical component be \(Y\).

For the horizontal component:

\(X = 30 - 34 \times \frac{8}{17} - 26 \times \frac{5}{13} = 4\)

For the vertical component:

\(Y = 34 \times \frac{15}{17} - 26 \times \frac{12}{13} = 6\)

Now, calculate the magnitude of the resultant force \(R\):

\(R = \sqrt{X^2 + Y^2} = \sqrt{4^2 + 6^2} = \sqrt{52} = 2\sqrt{13} = 7.21 \text{ N}\)

To find the direction \(\beta\) of the resultant force, use:

\(\beta = \arctan \left( \frac{Y}{X} \right) = \arctan \left( \frac{6}{4} \right) = 56.3^\circ\)

The direction is 56.3° above the 30 N force or anticlockwise from the 30 N force.

(a) Resolve forces horizontally:

\(20 \cos 30 = 25 \cos 60 + 10 \cos \alpha\)

\(17.32 = 12.5 + 10 \cos \alpha\)

\(\cos \alpha = 0.4821\)

\(\alpha = 61.2\)

Resolve forces vertically:

Resultant = \(20 \sin 30 + 10 \sin 61.2 - 25 \sin 60\)

= \(10 + 8.761 - 21.651\)

\(Magnitude of resultant force = 2.89 N\)

(b) Resolve forces horizontally and vertically:

\(X = 25 \cos 60 + 10 \cos 45 - 20 \cos 30\)

= \(12.5 + 7.07107 - 17.32051 = 2.25056\)

\(Y = 20 \sin 30 + 10 \sin 45 - 25 \sin 60\)

= \(10 + 7.07107 - 21.65064 = -4.57957\)

Resultant \(R = \sqrt{X^2 + Y^2}\)

= \(\sqrt{2.25056^2 + (-4.57957)^2} = 5.10 \text{ N}\)

Direction \(\alpha = \arctan \left( \frac{Y}{X} \right)\)

= \(\arctan \left( \frac{-4.57957}{2.25056} \right) = 63.8^\circ\) below positive x-axis

The forces acting at point A are 100 N to the left, and two 50 N forces at angles \(\alpha\) to the horizontal. The horizontal component of each 50 N force is \(50 \cos \alpha\).

The total horizontal component of the two 50 N forces is \(2 \times 50 \cos \alpha = 100 \cos \alpha\).

Given \(\cos \alpha = \frac{4}{5}\), the total horizontal component becomes \(100 \times \frac{4}{5} = 80\) N.

The resultant force is the difference between the 100 N force and the total horizontal component of the two 50 N forces:

\(100 - 80 = 20 \text{ N}\)

The direction of the resultant force is to the left.