Past Exam Questions

(a) To find the magnitude of the force in each of the struts AD and BD, resolve vertically:

\(500 + T \cos 45^{\circ} + T \cos 45^{\circ} - 100g = 0\)

Since \(T_A = T_B = T\), we have:

\(500 + 2T \cos 45^{\circ} = 100g\)

Solving for \(T\):

\(T \cos 45^{\circ} = \frac{100g - 500}{2}\)

\(T = \frac{500}{\sqrt{2}} = 354 \text{ N}\)

(b) To find the value of \(F\) for which the magnitude of the force in the strut AD is zero, resolve horizontally and vertically:

Vertically:

\(T_B \cos 45^{\circ} + 500 - 100g = 0\)

Horizontally:

\(F - T_B \sin 45^{\circ} = 0\)

Solving for \(F\):

\(F = T_B \sin 45^{\circ}\)

Since \(T_B \cos 45^{\circ} = 100g - 500\),

\(T_B = \frac{100g - 500}{\cos 45^{\circ}}\)

\(F = \frac{100g - 500}{\cos 45^{\circ}} \sin 45^{\circ} = 500 \text{ N}\)

Let the tension in the string attached to the wall be \(T\).

Resolve forces horizontally:

\(T \cos \alpha = 4 \sin 60^{\circ}\)

Resolve forces vertically:

\(T \sin \alpha + 4 \cos 60^{\circ} = 0.3g\)

Substitute \(g = 9.8 \, \text{m/s}^2\):

\(T \sin \alpha + 4 \times 0.5 = 0.3 \times 9.8\)

\(T \sin \alpha = 2.94 - 2\)

\(T \sin \alpha = 0.94\)

Using the horizontal equation:

\(T \cos \alpha = 4 \times \frac{\sqrt{3}}{2}\)

\(T \cos \alpha = 2\sqrt{3}\)

Divide the two equations:

\(\tan \alpha = \frac{0.94}{2\sqrt{3}}\)

\(\alpha = \arctan \left( \frac{0.94}{2\sqrt{3}} \right) \approx 76.9^{\circ}\)

Substitute \(\alpha\) back to find \(T\):

\(T = \frac{2\sqrt{3}}{\cos 76.9^{\circ}} \approx 2.05 \text{ N}\)

To solve for the tensions in the strings, we resolve the forces horizontally and vertically.

Let the tensions in the strings be \(T_1\) and \(T_2\).

Resolving horizontally:

\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)

Resolving vertically:

\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8g\)

Substituting \(g = 9.8 \text{ m/s}^2\), we have:

\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)

\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8 \times 9.8\)

Solving these equations simultaneously gives:

\(T_1 = 58.6 \text{ N}\)

\(T_2 = 41.4 \text{ N}\)

An alternative method using Lami's Theorem:

\(\frac{T_1}{\sin 135^{\circ}} = \frac{T_2}{\sin 150^{\circ}} = \frac{80}{\sin 75^{\circ}}\)

Solving for \(T_1\) and \(T_2\) gives the same results:

\(T_1 = 58.6 \text{ N}\)

\(T_2 = 41.4 \text{ N}\)

(a) Resolve horizontally: \(100 - T_U \sin 60 - T_L \sin 60 = 0\)

Resolve vertically: \(T_U \cos 60 - T_L \cos 60 - 5g = 0\)

Perpendicular to \(T_U\): \(T_L \cos 30 + 5g \cos 30 = 100 \cos 60\)

Solve for \(T_L\): \(T_L = 7.74 \text{ N}\)

(b) Resolve horizontally: \(X - T_{up} \sin 60 = 0\)

Resolve vertically: \(T_{up} \cos 60 - 5g = 0\)

Perpendicular to \(T_{up}\): \(5g \cos 30 = X \cos 60\)

Eliminate \(T_{up}\) and solve for \(X\): Least value of \(X = 86.6\)

To find the tensions in the strings, resolve the forces horizontally and vertically.

Resolving horizontally:

\(T_P \cos 60^{\circ} = T_R \cos 30^{\circ}\)

Resolving vertically:

\(T_P \sin 60^{\circ} = T_R \sin 30^{\circ} + 0.2g\)

Using \(g = 9.8 \text{ m/s}^2\), solve these equations simultaneously.

From the horizontal resolution:

\(T_P \cdot 0.5 = T_R \cdot \frac{\sqrt{3}}{2}\)

\(T_P = T_R \cdot \sqrt{3}\)

Substitute \(T_P = T_R \cdot \sqrt{3}\) into the vertical resolution:

\(T_R \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2} = T_R \cdot 0.5 + 0.2 \cdot 9.8\)

\(\frac{3}{2} T_R = 0.5 T_R + 1.96\)

\(T_R = 2 \text{ N}\)

Substitute \(T_R = 2 \text{ N}\) back to find \(T_P\):

\(T_P = 2 \cdot \sqrt{3} \approx 3.46 \text{ N}\)

Resolve the forces horizontally:

\(20 \cos 60^\circ = T \cos 45^\circ\)

\(T = \frac{20 \cos 60^\circ}{\cos 45^\circ} = 10\sqrt{2}\) or 14.1 \text{ N}

Resolve the forces vertically:

\(20 \sin 60^\circ + T \sin 45^\circ = mg\)

\(m = \frac{20 \sin 60^\circ + T \sin 45^\circ}{g}\)

Substitute \(T = 10\sqrt{2}\):

\(m = \frac{20 \times \frac{\sqrt{3}}{2} + 10\sqrt{2} \times \frac{\sqrt{2}}{2}}{g} = 2.73 \text{ kg}\)

To solve for the tensions in the strings, we resolve the forces vertically and horizontally.

Vertically, the equilibrium condition gives:

\(T_A \times \frac{4}{5} + T_B \times \frac{3}{5} = 5\)

Horizontally, the equilibrium condition gives:

\(T_A \times \frac{3}{5} = T_B \times \frac{4}{5}\)

Solving these equations simultaneously, we find:

\(T_A = 1.6 \text{ N}\)

\(T_B = 1.2 \text{ N}\)

To solve for the tension \(T\) and the mass \(m\), we resolve forces horizontally and vertically.

Horizontally, the equilibrium condition gives:

\(T \sin 30^{\circ} + T \sin 40^{\circ} - 2 = 0\)

Vertically, the equilibrium condition gives:

\(T \cos 30^{\circ} - T \cos 40^{\circ} - mg = 0\)

Solving the horizontal equation for \(T\):

\(T (\sin 30^{\circ} + \sin 40^{\circ}) = 2\)

\(T = \frac{2}{\sin 30^{\circ} + \sin 40^{\circ}}\)

Substituting the values of \(\sin 30^{\circ} = 0.5\) and \(\sin 40^{\circ} \approx 0.6428\):

\(T = \frac{2}{0.5 + 0.6428} \approx 1.75\)

Substitute \(T = 1.75\) into the vertical equation:

\(1.75 \cos 30^{\circ} - 1.75 \cos 40^{\circ} = mg\)

\(mg = 1.75 (\cos 30^{\circ} - \cos 40^{\circ})\)

Using \(\cos 30^{\circ} \approx 0.866\) and \(\cos 40^{\circ} \approx 0.766\):

\(mg = 1.75 (0.866 - 0.766)\)

\(mg = 1.75 \times 0.1 = 0.175\)

\(m = \frac{0.175}{g}\)

Assuming \(g = 9.8\):

\(m \approx \frac{0.175}{9.8} \approx 0.0175\)

To solve for the tension in the string and the value of X, we resolve the forces acting on the ring R.

1. Resolve vertically: \(X \sin 60^{\circ} + T \sin 30^{\circ} - 0.2g = 0\)

2. Resolve horizontally: \(X \cos 60^{\circ} - T - T \cos 30^{\circ} = 0\)

Solving these equations simultaneously:

From the vertical resolution, substitute the known values: \(X \cdot \frac{\sqrt{3}}{2} + T \cdot \frac{1}{2} = 0.2 \cdot 9.8\)

From the horizontal resolution: \(X \cdot \frac{1}{2} = T + T \cdot \frac{\sqrt{3}}{2}\)

Solving these equations gives: \(X = 2\) \(T = 0.536 \text{ N}\)

To solve this problem, we need to resolve the forces acting on the ring R both vertically and horizontally.

Vertically: The vertical components of the tension must balance the weight of the ring. The weight of the ring is given by:

\(0.2g\)

where \(g\) is the acceleration due to gravity (approximately 9.8 m/s²). The vertical components of the tension are:

\(T \sin 70^{\circ} + T \sin 45^{\circ} = 0.2g\)

Solving for \(T\):

\(T(\sin 70^{\circ} + \sin 45^{\circ}) = 0.2 \times 9.8\)

\(T = \frac{0.2 \times 9.8}{\sin 70^{\circ} + \sin 45^{\circ}} \approx 1.21 \text{ N}\)

Horizontally: The horizontal components of the tension must balance the horizontal force \(P\):

\(P + T \cos 70^{\circ} = T \cos 45^{\circ}\)

Substituting the value of \(T\):

\(P + 1.21 \cos 70^{\circ} = 1.21 \cos 45^{\circ}\)

Solving for \(P\):

\(P = 1.21 \cos 45^{\circ} - 1.21 \cos 70^{\circ} \approx 0.443 \text{ N}\)

To solve this problem, we resolve the forces acting on the ring R in both horizontal and vertical directions.

1. Resolve horizontally: The horizontal component of the tension T in the string is balanced by the horizontal component of the 2.5 N force.

\(T \cos 45^{\circ} + T \cos 45^{\circ} = 2.5 \cos 45^{\circ}\)

\(2T \cos 45^{\circ} = 2.5 \cos 45^{\circ}\)

\(T = 1.25 \text{ N}\)

2. Resolve vertically: The vertical component of the tension balances the weight of the ring.

\(2.5 \sin 45^{\circ} = mg\)

\(m = \frac{2.5 \sin 45^{\circ}}{g}\)

Using \(g = 9.8 \text{ m/s}^2\),

\(m = \frac{2.5 \times \frac{\sqrt{2}}{2}}{9.8} \approx 0.177 \text{ kg}\)

Thus, the tension in the string is 1.25 N and the mass of the ring is 0.177 kg.

To solve the problem, we resolve the forces acting on the ring R horizontally and vertically.

1. Resolving horizontally:

\(T \cos \theta = 11.2\)

2. Resolving vertically:

\(T \sin \theta = 0.16g\)

where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).

Substituting \(g = 9.8\), we have:

\(T \sin \theta = 0.16 \times 9.8 = 1.568\)

Now, using the given mark-scheme:

\(T \cos \theta = 4.8\)

\(T \sin \theta = 6.4\)

We can find \(T\) and \(\theta\) using these equations:

\(T^2 = 4.8^2 + 6.4^2\)

\(T^2 = 23.04 + 40.96 = 64\)

\(T = \sqrt{64} = 8\)

To find \(\theta\):

\(\tan \theta = \frac{6.4}{4.8}\)

\(\theta = \arctan \left( \frac{6.4}{4.8} \right) \approx 53.1^{\circ}\)

To solve the problem, we resolve the forces acting on the ring R horizontally and vertically.

Horizontally, the forces are balanced as:

\(T \cos \theta + T \sin \theta = 15.5\)

Vertically, the forces are balanced as:

\(-T \cos \theta + T \sin \theta = 8.5\)

Solving these two equations simultaneously, we find:

\(T \sin \theta = 12\)

\(T \cos \theta = 3.5\)

To find \(\theta\), we use the identity:

\(\tan \theta = \frac{T \sin \theta}{T \cos \theta} = \frac{12}{3.5}\)

\(\tan \theta = \frac{12}{3.5} = 3.4286\)

Thus, \(\theta = \arctan(3.4286) \approx 73.7^\circ\) or 1.29 radians.

To solve this problem, we need to resolve the forces acting on the ring R both vertically and horizontally.

(i) Resolving vertically:

The vertical components of the tension must balance the weight of the ring. Let T be the tension in the string.

From the diagram, the vertical components are:

\(T \sin 50^\circ\) and \(T \sin 20^\circ\)

These must sum to the weight of the ring:

\(T \sin 50^\circ = T \sin 20^\circ + 0.8g\)

Solving for T using \(g = 9.8 \text{ m/s}^2\):

\(T \sin 50^\circ - T \sin 20^\circ = 0.8 \times 9.8\)

\(T(\sin 50^\circ - \sin 20^\circ) = 7.84\)

\(T = \frac{7.84}{\sin 50^\circ - \sin 20^\circ}\)

Calculating gives \(T \approx 18.9 \text{ N}\).

(ii) Resolving horizontally:

The horizontal components of the tension must balance the horizontal force X.

\(X = T \cos 50^\circ + T \cos 20^\circ\)

Substitute \(T = 18.9 \text{ N}\):

\(X = 18.9 \cos 50^\circ + 18.9 \cos 20^\circ\)

Calculating gives \(X \approx 29.9 \text{ N}\).

(a) To show \(\left( \frac{14.4}{30 - P} \right)^2 + \left( \frac{28.8}{P + 30} \right)^2 = 1\), resolve forces horizontally and vertically.

Horizontal: \((33 + 15) \times \frac{3}{5} = P \cos \theta + 30 \cos \theta\)

Vertical: \(15 \times \frac{4}{5} + 30 \sin \theta = 33 \times \frac{4}{5} + P \sin \theta\)

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

\(\cos \theta = \frac{28.8}{P + 30}\) and \(\sin \theta = \frac{14.4}{30 - P}\)

Substitute these into \(\cos^2 \theta + \sin^2 \theta = 1\) to get:

\(\left( \frac{14.4}{30 - P} \right)^2 + \left( \frac{28.8}{P + 30} \right)^2 = 1\)

(b) Substitute \(P = 6\):

\(\left( \frac{14.4}{24} \right)^2 + \left( \frac{28.8}{36} \right)^2 = \left( \frac{3}{5} \right)^2 + \left( \frac{4}{5} \right)^2 = 0.36 + 0.64 = 1\)

Thus, \(\theta = 36.9^\circ\).

To show that the forces are in equilibrium, we need to resolve the forces horizontally and vertically and show that the net force in each direction is zero.

Resolving horizontally:

\(X = 78 \times \frac{5}{13} - 50 \times \frac{3}{5} = 78 \cos 67.4^\circ - 50 \cos 53.1^\circ\)

\(X = 30 - 30 = 0\)

Resolving vertically:

\(Y = 78 \times \frac{12}{13} + 50 \times \frac{4}{5} - 112 = 78 \sin 67.4^\circ + 50 \sin 53.1^\circ - 112\)

\(Y = 72 + 40 - 112 = 0\)

Since both \(X = 0\) and \(Y = 0\), the forces are in equilibrium.

Alternatively, using Lami's theorem:

\(\frac{112}{\sin 59.5^\circ} = \frac{50}{\sin 157.4^\circ} = \frac{78}{\sin 143.1^\circ}\)

Calculating the exact values:

\(\frac{112}{56/65} = \frac{50}{5/13} = \frac{78}{3/5} = 130\)

Thus, the forces satisfy Lami's theorem, confirming equilibrium.

Since the forces are in equilibrium, the sum of the horizontal and vertical components must be zero.

Resolve horizontally:

\(F \cos \alpha = 15 \cos 20^\circ - 5\)

\(F \cos \alpha = 9.095\)

Resolve vertically:

\(F \sin \alpha = 15 \sin 20^\circ + 25\)

\(F \sin \alpha = 30.13\)

Using Pythagoras' theorem to find \(F\):

\(F = \sqrt{(15 \cos 20^\circ - 5)^2 + (15 \sin 20^\circ + 25)^2}\)

\(F = 31.5\)

Using trigonometry to find \(\alpha\):

\(\alpha = \arctan \left( \frac{15 \sin 20^\circ + 25}{15 \cos 20^\circ - 5} \right)\)

\(\alpha = 73.2^\circ\)

To find the single additional force that will produce equilibrium, we need to resolve the given forces into their components and find the resultant force. The additional force will be equal in magnitude and opposite in direction to this resultant force.

Resolve the forces into horizontal (x-axis) and vertical (y-axis) components:

Horizontal component, \(X = 18 + 12\sin 60^\circ - 8\sin 30^\circ = 14 + 6\sqrt{3}\).

Vertical component, \(Y = 8\cos 30^\circ + 12\cos 60^\circ = 6 + 4\sqrt{3}\).

The magnitude of the resultant force is given by:

\(\sqrt{(14 + 6\sqrt{3})^2 + (6 + 4\sqrt{3})^2} = 27.6 \text{ N}\)

The direction of the resultant force is given by:

\(\arctan \left( \frac{6 + 4\sqrt{3}}{14 + 6\sqrt{3}} \right) = 27.9^\circ\)

Thus, the additional force required for equilibrium has a magnitude of 27.6 N and is directed 27.9° below the negative x-axis.

To solve for \(\theta\) and \(P\), we resolve the forces horizontally and vertically, using the fact that the forces are in equilibrium.

Resolving horizontally:

\(3 \cos 60^{\circ} = 2 \cos \theta\)

\(\frac{3}{2} = 2 \cos \theta\)

\(\cos \theta = \frac{3}{4}\)

\(\theta = \cos^{-1}\left(\frac{3}{4}\right) \approx 41.4^{\circ}\)

Resolving vertically:

\(P = 3 \sin 60^{\circ} + 2 \sin \theta\)

\(P = 3 \times \frac{\sqrt{3}}{2} + 2 \times \sin 41.4^{\circ}\)

\(P \approx 3.92 \text{ N}\)

Since the forces are in equilibrium, we can use the cosine rule and Lami's theorem to find the angles between the forces.

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

\(10 \cos \alpha = 8 \quad \text{or} \quad 10 \cos \beta = 6\)

Solving for \(\alpha\) and \(\beta\):

\(\alpha = 36.9^{\circ} \quad \text{or} \quad \beta = 53.1^{\circ}\)

Using Lami's theorem:

\(\frac{10}{\sin 90^{\circ}} = \frac{6}{\sin \gamma} = \frac{8}{\sin \delta}\)

Solving for \(\gamma\) and \(\delta\):

\(\gamma = 143.1^{\circ}\) (angle between 8 N and 10 N)

\(\delta = 126.9^{\circ}\) (angle between 6 N and 10 N)