Past Exam Questions

To find \(\theta\) and \(P\), we resolve the forces horizontally and vertically, and use Lami's theorem.

Step 1: Resolve Vertically

Using vertical equilibrium: \(2P \sin \theta = P \sin 60\)

\(\Rightarrow 2 \sin \theta = \sin 60\)

\(\Rightarrow \theta = 25.7^\circ\)

Step 2: Resolve Horizontally

Using horizontal equilibrium: \(2P \cos \theta + P \cos 60 = 10\)

Substitute \(\theta = 25.7^\circ\):

\(2P \cos 25.7 + P \cdot 0.5 = 10\)

\(\Rightarrow P = 4.34 \text{ N}\)

Alternative Method: Lami's Theorem

Using Lami's theorem:

\(\frac{2P}{\sin 120} = \frac{P}{\sin(180 - \theta)} = \frac{10}{\sin(60 + \theta)}\)

Solve for \(\theta\) and \(P\) using the above relationships, confirming \(\theta = 25.7^\circ\) and \(P = 4.34 \text{ N}\).

To find the values of P and θ, we resolve the forces vertically and horizontally, and use Lami's Theorem.

Vertical Resolution:
The vertical components of the forces must sum to zero for equilibrium:

\(3P \sin 55^{\circ} + P \sin θ + P \sin θ = 20\)

\(3P \sin 55^{\circ} = 20\)

Solving for P:

\(P = \frac{20}{3 \sin 55^{\circ}} \approx 8.14 \text{ N}\)

Horizontal Resolution:
The horizontal components of the forces must also sum to zero:

\(3P \cos 55^{\circ} = 2P \cos θ\)

\(\cos θ = \frac{3 \cos 55^{\circ}}{2}\)

Solving for θ:

\(\cos θ = 1.5 \cos 55^{\circ}\)

\(θ \approx 30.6^{\circ}\)

Alternatively, using Lami's Theorem:

\(\frac{3P}{\sin 90^{\circ}} = \frac{20}{\sin 125^{\circ}}\)

Solving gives:

\(P = 8.14 \text{ N}\)

And for the horizontal forces:

\(\frac{3P}{\sin 90^{\circ}} = \frac{2P \cos θ}{\sin 145^{\circ}}\)

\(\cos θ = 1.5 \sin 145^{\circ}\)

\(θ \approx 30.6^{\circ}\)

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

First, resolve horizontally:

\(2F + F \cos 60^\circ = 15 \cos \alpha\)

Next, resolve vertically:

\(F \sin 60^\circ = 15 \sin \alpha\)

Using these two equations, we can solve for \(F\) and \(\alpha\). From the mark scheme, we have:

\(F = 5.67\) and \(\alpha = 19.1\)

Alternatively, \(F = \frac{15 \sqrt{7}}{7}\).

To solve for \(P\) and \(\theta\), we resolve the forces horizontally and vertically.

For horizontal equilibrium:

\(12\cos75^\circ + P\cos\theta^\circ = 18\cos65^\circ\)

For vertical equilibrium:

\(18\sin65^\circ + 12\sin75^\circ = 15 + P\sin\theta^\circ\)

We can solve these simultaneous equations by eliminating either \(\theta\) or \(P\).

Using the given mark scheme:

\(P^2 = (18\sin65^\circ + 12\sin75^\circ - 15)^2 + (18\cos65^\circ - 12\cos75^\circ)^2\)

or

\(\theta = \arctan\left(\frac{18\sin65^\circ + 12\sin75^\circ - 15}{18\cos65^\circ - 12\cos75^\circ}\right)\)

Solving these gives:

\(P = 13.7\) or \(\theta = 70.8\)

To solve for \(P\) and \(\theta\), we resolve the forces horizontally and vertically.

Given \(\tan \alpha = \frac{7}{24}\), we have \(\cos \alpha = \frac{24}{25}\) and \(\sin \alpha = \frac{7}{25}\).

Resolving horizontally:

\(P \cos \theta = 48 \cos \alpha - 14 \sin \alpha\)

\(P \cos \theta = 48 \left(\frac{24}{25}\right) - 14 \left(\frac{7}{25}\right) = 42.16\)

Resolving vertically:

\(P \sin \theta = 50 - 48 \sin \alpha - 14 \cos \alpha\)

\(P \sin \theta = 50 - 48 \left(\frac{7}{25}\right) - 14 \left(\frac{24}{25}\right) = 23.12\)

Using Pythagoras' theorem:

\(P = \sqrt{42.16^2 + 23.12^2} = 48.1\)

To find \(\theta\):

\(\tan \theta = \frac{23.12}{42.16}\)

\(\theta = 28.7^\circ\)

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

Resolving horizontally:

\(P \cos 25^\circ = 22 + 16 \cos 55^\circ\)

Resolving vertically:

\(Q + 16 \sin 55^\circ = P \sin 25^\circ\)

First, solve the horizontal equation for \(P\):

\(P = \frac{22 + 16 \cos 55^\circ}{\cos 25^\circ}\)

Substitute \(P\) into the vertical equation:

\(Q + 16 \sin 55^\circ = \left(\frac{22 + 16 \cos 55^\circ}{\cos 25^\circ}\right) \sin 25^\circ\)

Solve for \(Q\):

\(Q = \left(\frac{22 + 16 \cos 55^\circ}{\cos 25^\circ}\right) \sin 25^\circ - 16 \sin 55^\circ\)

Calculating these values gives:

\(P \approx 34.4\)

\(Q \approx 1.43\)

(i) To find F, resolve the forces in the horizontal direction. The equilibrium condition gives:

\(15 + F \cos 60^\circ = F \cos 30^\circ\)

Using \(\cos 60^\circ = \frac{1}{2}\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), the equation becomes:

\(15 + \frac{F}{2} = \frac{F\sqrt{3}}{2}\)

Rearranging gives:

\(15 = \frac{F\sqrt{3}}{2} - \frac{F}{2}\)

\(15 = \frac{F(\sqrt{3} - 1)}{2}\)

Solving for \(F\):

\(F = \frac{30}{\sqrt{3} - 1}\)

Rationalizing the denominator:

\(F = \frac{30(\sqrt{3} + 1)}{2} = 15(\sqrt{3} + 1)\)

Calculating gives \(F \approx 41.0\).

(ii) To find G, resolve the forces in the vertical direction. The equilibrium condition gives:

\(G = F(\sin 30^\circ + \sin 60^\circ)\)

Using \(\sin 30^\circ = \frac{1}{2}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), the equation becomes:

\(G = F\left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right)\)

\(G = F\left(\frac{1 + \sqrt{3}}{2}\right)\)

Substituting \(F = 41.0\):

\(G = 41.0 \times \frac{1 + \sqrt{3}}{2}\)

Calculating gives \(G \approx 56.0\).

Resolve the forces into their components. For the force of 25 N, the angle with the negative x-axis is tan^-1 0.75, which gives \cos \theta = 0.8 and \sin \theta = 0.6.

The components of the force F are:

\(F_x = F \cos \theta = 25 \times 0.8 = 20\)

\(F_y = F \sin \theta = 63 - 25 \times 0.6 = 48\)

Using equilibrium conditions, \(F = \sqrt{F_x^2 + F_y^2}\) or \(\tan \theta = \frac{F_y}{F_x}\).

Thus, \(F = 52 \, \text{N}\) and \(\tan \theta = 2.4\).

To solve for \(\alpha\) and \(F\), we use the equilibrium condition for forces. The forces are in equilibrium, so the sum of the forces in any direction is zero.

Resolve the forces horizontally and vertically:

Horizontal: \(F \cos \alpha = 12\)

Vertical: \(F \sin \alpha + 12 \sin \alpha = 15\)

Using the horizontal equation, \(F = \frac{12}{\cos \alpha}\).

Substitute into the vertical equation:

\(\frac{12 \sin \alpha}{\cos \alpha} + 12 \sin \alpha = 15\)

\(12 \tan \alpha + 12 \sin \alpha = 15\)

\(12 \sin \alpha (1 + \cos \alpha) = 15\)

\(\sin \alpha = \frac{12}{15}\)

\(\alpha = \arcsin \left( \frac{12}{15} \right) = 53.1^\circ\)

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

\(F = \frac{12}{\cos 53.1^\circ} = 9 \text{ N}\)

To solve for \(\alpha\) and \(F\), we resolve the forces in the horizontal and vertical directions.

Resolving horizontally:

\(6 \cos \alpha + 5 \cos (90^\circ - \alpha) = F\)

Resolving vertically:

\(6 \sin \alpha - 5 \sin (90^\circ - \alpha) = F\)

Using the identity \(\cos(90^\circ - \alpha) = \sin \alpha\) and \(\sin(90^\circ - \alpha) = \cos \alpha\), we have:

\(6 \cos \alpha + 5 \sin \alpha = 6 \sin \alpha - 5 \cos \alpha\)

Rearranging gives:

\(11 \cos \alpha = \sin \alpha\)

\(\tan \alpha = \frac{1}{11}\)

Solving for \(\alpha\), we find:

\(\alpha = 84.8^\circ\)

Substituting \(\alpha = 84.8^\circ\) back into the horizontal resolution equation:

\(F = 6 \cos 84.8^\circ + 5 \sin 84.8^\circ\)

\(F = 5.52 \text{ N}\)

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

Resolve the forces into horizontal and vertical components:

• Horizontal: \(F \cos \theta = 10\)
• Vertical: \(F \sin \theta = 13\)

Using the tangent function, we have:

\(\tan \theta = \frac{13}{10}\)

Thus, \(\theta = \arctan \left( \frac{13}{10} \right) \approx 52.4^\circ\).

Using Pythagoras' theorem for the resultant force:

\(F^2 = 10^2 + 13^2\)

\(F = \sqrt{10^2 + 13^2} = \sqrt{269} \approx 16.4 \text{ N}\)

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

Resolving forces in the vertical ('j') direction:

\(F \sin 50^\circ = F \sin 20^\circ + 12\)

Solving for \(F\):

\(F (\sin 50^\circ - \sin 20^\circ) = 12\)

\(F = \frac{12}{\sin 50^\circ - \sin 20^\circ} \approx 28.3\)

Resolving forces in the horizontal ('i') direction:

\(G = F \cos 50^\circ + F \cos 20^\circ\)

Substitute \(F = 28.3\):

\(G = 28.3 \cos 50^\circ + 28.3 \cos 20^\circ\)

\(G \approx 44.8\)

To solve for \(F\) and \(\alpha\), we use the equilibrium condition, which states that the sum of the forces in any direction must be zero.

Resolve horizontally:

\(F \cos \alpha - 20 \cos 40 - 100 \sin 20 = 0\)

\(F \cos \alpha = 49.5229 \ldots\)

Resolve vertically:

\(F \sin \alpha + 20 \sin 40 - 100 \cos 20 = 0\)

\(F \sin \alpha = 81.1135 \ldots\)

Now, solve for \(F\):

\(F = \sqrt{(49.5229 \ldots)^2 + (81.1135 \ldots)^2}\)

\(F = 95.0\)

To find \(\alpha\):

\(\tan \alpha = \frac{81.1135 \ldots}{49.5229 \ldots}\)

\(\alpha = 58.6\)

Since the system is in equilibrium, the sum of the forces in any direction is zero. We resolve the forces horizontally and vertically.

For horizontal components:

\(P \cos \theta = (36 - 24) \cos 36.9^{\circ}\)

\(P \cos \theta = 9.6\)

For vertical components:

\(P \sin \theta + 20 = (24 + 36) \sin 36.9^{\circ}\)

\(P \sin \theta + 20 = 36\)

\(P \sin \theta = 16\)

Using the equations:

\(P \cos \theta = 9.6\)

\(P \sin \theta = 16\)

We find \(P\) using:

\(P = \sqrt{16^2 + 9.6^2}\)

\(P = 18.7\)

To find \(\theta\):

\(\theta = \arctan \left( \frac{16}{9.6} \right)\)

\(\theta = 59.0^{\circ}\)

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

For the vertical direction:

\(F \sin \theta + 20 \sin 60^\circ - 30 \sin \alpha - 40 \sin \beta = 0\)

Substituting the given values:

\(F \sin \theta + 20 \times 0.866 - 30 \times 0.28 - 40 \times 0.6 = 0\)

\(F \sin \theta = 15.07949 \ldots\)

For the horizontal direction:

\(F \cos \theta + 40 \times 0.8 - 30 \times 0.96 - 20 \cos 60^\circ = 0\)

\(F \cos \theta = 6.8\)

Now, find \(\theta\):

\(\theta = \arctan \left( \frac{15.0794 \ldots}{6.8} \right)\)

\(\theta = 65.7^\circ\)

Finally, find \(F\):

\(F = \sqrt{15.07949^2 + 6.8^2}\)

\(F = 16.5\)

Resolve forces horizontally:

\(P \cos \theta = 12 + 8 \cos 30^\circ - 10 \cos 45^\circ\)

\(P \cos \theta = 11.857\)

Resolve forces vertically:

\(P \sin \theta = 10 \sin 45^\circ - 8 \sin 30^\circ\)

\(P \sin \theta = 3.071\)

Calculate \(P\):

\(P = \sqrt{11.857^2 + 3.071^2}\)

\(P = 12.2\)

Calculate \(\theta\):

\(\theta = \arctan \left( \frac{3.071}{11.857} \right)\)

\(\theta = 14.5\)

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

Resolving horizontally:

\(F \cos \alpha = 40 \sin 30^\circ + 20 \sin 60^\circ - 50 \sin 45^\circ\)

\(F \cos \alpha = 1.965 \ldots\)

Resolving vertically:

\(F \sin \alpha = 50 \cos 45^\circ + 20 \cos 60^\circ - 40 \cos 30^\circ\)

\(F \sin \alpha = 10.714 \ldots\)

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

\(F = \sqrt{(1.965 \ldots)^2 + (10.714 \ldots)^2} = 10.9 \text{ N}\)

To find \(\alpha\):

\(\alpha = \arctan\left(\frac{10.714 \ldots}{1.965 \ldots}\right) = 79.6^\circ\)

Resolving forces horizontally, we have:

\(20 \cos \theta = 4P \cos 30\)

\(20 \cos \theta = 4P \times \frac{\sqrt{3}}{2}\)

\(\cos \theta = \frac{\sqrt{3}}{10} P\)

Resolving forces vertically, we have:

\(4P + 2P \sin 30 = 20 \sin \theta\)

\(4P + P = 20 \sin \theta\)

\(\sin \theta = \frac{P}{4}\)

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

\(\left( \frac{\sqrt{3}}{10} P \right)^2 + \left( \frac{P}{4} \right)^2 = 1\)

\(\frac{3}{100} P^2 + \frac{1}{16} P^2 = 1\)

Solving for \(P\), we find \(P = 3.29\).

Substituting \(P\) back to find \(\theta\):

\(\cos \theta = \frac{\sqrt{3}}{10} \times 3.29\)

\(\theta = 55.3\)

To solve for P and \(\theta\), we resolve the forces horizontally and vertically, as the system is in equilibrium.

Resolving horizontally:

\(P \cos \theta = 32 \cos 20^\circ - 17 \sin 55^\circ\)

\(P \cos \theta = 16.1446\)

Resolving vertically:

\(P \sin \theta = 40 + 17 \cos 55^\circ - 32 \sin 20^\circ\)

\(P \sin \theta = 38.8062\)

Using Pythagoras' theorem to find P:

\(P = \sqrt{(17 \cos 55^\circ - 32 \sin 20^\circ + 40)^2 + (32 \cos 20^\circ - 17 \cos 35^\circ)^2}\)

\(P = 42.0\)

Using trigonometry to find \(\theta\):

\(\theta = \arctan \left( \frac{17 \cos 55^\circ - 32 \sin 20^\circ + 40}{32 \cos 20^\circ - 17 \cos 35^\circ} \right)\)

\(\theta = 67.4\)

(a) Resolving forces vertically and horizontally:

Vertical: \(F \sin \theta + 40 \sin 60^\circ - 50 = 0\)

Horizontal: \(F \cos \theta + 10 - 40 \cos 60^\circ = 0\)

From horizontal resolution: \(F \cos \theta = 10\)

From vertical resolution: \(F \sin \theta = 50 - 20\sqrt{3}\)

\(\theta = \arctan(5 - 2\sqrt{3}) \approx 56.9^\circ\)

\(F = \sqrt{(50 - 20\sqrt{3})^2 + 10^2} \approx 18.3 \text{ N}\)

(b) Given \(F = 10\sqrt{2}\) and \(\theta = 45^\circ\):

Vertical component: \(Y = 10\sqrt{2} \sin 45^\circ + 40 \sin 60^\circ - 50 = 20\sqrt{3} - 40\)

Horizontal component: \(X = 10\sqrt{2} \cos 45^\circ + 10 - 40 \cos 60^\circ = 0\)

Resultant force is \(40 - 20\sqrt{3} \text{ N}\) in the same direction as the \(50 \text{ N}\) force.