Past Exam Questions

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.

(i) Resolve the forces horizontally and vertically:

Horizontal component: \(25 \cos 30^{\circ} - 15 \cos 40^{\circ} = 10.1599 \ldots\)

Vertical component: \(25 \sin 30^{\circ} + 15 \sin 40^{\circ} - 30 = -7.8581 \ldots\)

Magnitude of the resultant force:

\(\sqrt{(10.1599)^2 + (-7.8581)^2} = 12.8 \text{ N}\)

Direction: Angle 37.7° below the horizontal in the direction BA.

(ii) For zero component in the direction BC:

\(F \cos 40^{\circ} = 25 \cos 30^{\circ}\)

\(F = 28.3\)

New resultant force:

\(28.26 \sin 40^{\circ} + 25 \sin 30^{\circ} - 30 = 0.667 \text{ N upwards}\)

To find the value of \(F\), we need to consider the horizontal components of the forces, as the resultant is perpendicular to \(F\).

The horizontal component of the \(20 \text{ N}\) force is \(20 \cos 60^\circ\).

The horizontal component of the \(30 \text{ N}\) force is \(30 \cos 60^\circ\).

The horizontal component of the \(F \text{ N}\) force is \(-F\) (since it acts to the left).

Since the resultant is perpendicular to \(F\), the sum of the horizontal components must be zero:

\(20 \cos 60^\circ + 30 \cos 60^\circ - F = 0\)

\(F = 20 \cos 60^\circ + 30 \cos 60^\circ\)

\(F = 20 \times \frac{1}{2} + 30 \times \frac{1}{2}\)

\(F = 10 + 15\)

\(F = 25\)

(i) Resolve the forces horizontally and vertically:

Horizontal component: \(X = 75 + 50 \cos 60^\circ = 100\)

Vertical component: \(Y = 50 \sin 60^\circ = 43.3\)

Resultant force: \(\sqrt{100^2 + 43.3^2} = 109 \text{ N}\)

Direction: \(\text{Angle} = \arctan\left(\frac{43.3}{100}\right) = 23.4^\circ\) anticlockwise from the positive x-axis.

(ii) For equilibrium, resolve forces:

Horizontal: \(50 \cos \alpha - F \cos 50^\circ = 0\)

Vertical: \(50 \sin \alpha - 3F - F \sin 50^\circ = 0\)

\(\tan \alpha = \frac{3F + F \sin 50^\circ}{F \cos 50^\circ}\)

Solving gives \(\alpha = 80.3^\circ\) and \(F = 13.1 \text{ N}\).

(i) To find the total force in the direction of the river, resolve each force into its horizontal component:

\(X = 60 \cos 25^{\circ} + 50 \cos 15^{\circ}\)

Calculating gives:

\(X = 54.49 + 48.30 = 102.79 \text{ N}\)

Rounded to 103 N.

(ii) To find the magnitude and direction of the resultant force, resolve each force into its vertical component:

\(Y = 60 \sin 25^{\circ} - 50 \sin 15^{\circ}\)

Calculating gives:

\(Y = 25.35 - 12.94 = 12.41 \text{ N}\)

Now, use Pythagoras' theorem to find the magnitude:

\(R = \sqrt{X^2 + Y^2} = \sqrt{102.79^2 + 12.41^2}\)

\(R = \sqrt{10568.38 + 154.08} = \sqrt{10722.46} = 103.4 \text{ N}\)

To find the direction, use:

\(\alpha = \arctan\left(\frac{Y}{X}\right) = \arctan\left(\frac{12.41}{102.79}\right)\)

\(\alpha = 6.9^{\circ}\)

To find the resultant force, resolve the forces horizontally and vertically.

For the horizontal component:

\(X = 7 - 8 \cos \alpha - 6 \sin \alpha\)

Given \(\sin \alpha = \frac{3}{5}\), use \(\cos \alpha = \frac{4}{5}\) (since \(\sin^2 \alpha + \cos^2 \alpha = 1\)).

\(X = 7 - 8 \times \frac{4}{5} - 6 \times \frac{3}{5} = -3\)

For the vertical component:

\(Y = 8 \sin \alpha - 6 \cos \alpha\)

\(Y = 8 \times \frac{3}{5} - 6 \times \frac{4}{5} = 0\)

The resultant force is \(\sqrt{X^2 + Y^2} = \sqrt{(-3)^2 + 0^2} = 3 \text{ N}\).

Since \(X = -3\), the direction is to the left.

(i) Resolve the forces horizontally and vertically:

Horizontal component: \(R_x = 40 \times \frac{24}{25} - 30 \times \frac{7}{25} = 30\)

Vertical component: \(R_y = 50 - 40 \times \frac{7}{25} - 30 \times \frac{24}{25} = 10\)

Magnitude of the resultant force:

\(R = \sqrt{R_x^2 + R_y^2} = \sqrt{30^2 + 10^2} = 31.6 \text{ N}\)

Direction of the resultant force:

\(\theta = \arctan \left( \frac{R_y}{R_x} \right) = \arctan \left( \frac{10}{30} \right) = 18.4^\circ\)

(ii) To make the resultant act in the positive \(x\)-direction, set \(R_y = 0\):

\(50 - P \times \frac{7}{25} - 30 \times \frac{24}{25} = 0\)

Solving for \(P\):

\(P = 40\)

Resolve the forces in the horizontal direction:

\(120 \cos 75^{\circ} = 150 - 100 - P \cos \theta\)

\(120 \cos 75^{\circ} = 50 - P \cos \theta\)

Resolve the forces in the vertical direction:

\(120 \sin 75^{\circ} = P \sin \theta\)

Using the Pythagorean identity for the resultant force:

\(P^2 = (P \cos \theta)^2 + (P \sin \theta)^2\)

\(P^2 = 14400 - 12000 \cos 75^{\circ} + 2500\)

Alternatively, use the tangent identity:

\(\tan \theta = \frac{120 \sin 75^{\circ}}{50 - 120 \cos 75^{\circ}}\)

Solving these equations gives:

\(P = 117\) N

\(\theta = 80.7^{\circ}\)

(i) To find the resultant of the forces in Fig. 1, we calculate the components:

The x-component is given by:

\(x = 4 + 8 \cos 30^\circ + 12 \cos 60^\circ = 10 + 4\sqrt{3} \approx 16.928\)

The y-component is given by:

\(y = 8 \sin 30^\circ + 12 \sin 60^\circ + 16 = 20 + 6\sqrt{3} \approx 30.392\)

The magnitude of the resultant force is:

\(R = \sqrt{x^2 + y^2} = \sqrt{16.928^2 + 30.392^2} \approx 34.8 \text{ N}\)

The direction \(\theta\) with the 4 N force is:

\(\theta = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{30.392}{16.928} \right) \approx 60.9^\circ\)

(ii) For Fig. 2, the forces exchange directions. The magnitude remains the same:

\(R = 34.8 \text{ N}\)

The direction \(\theta\) with the 16 N force is:

\(\theta = 90^\circ - 60.9^\circ = 29.1^\circ\)

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

For the x-direction:

\(X = 5 - 7\cos 60^\circ - 3\cos 30^\circ\)

\(X = 5 - 7 \times 0.5 - 3 \times \frac{\sqrt{3}}{2}\)

\(X = 5 - 3.5 - 2.598 \approx -1.098\)

For the y-direction:

\(Y = 7\sin 60^\circ - 3\sin 30^\circ - 4\)

\(Y = 7 \times \frac{\sqrt{3}}{2} - 3 \times 0.5 - 4\)

\(Y = 6.062 - 1.5 - 4 \approx 0.562\)

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

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

\(R = \sqrt{(-1.098)^2 + (0.562)^2}\)

\(R = \sqrt{1.206 + 0.316} \approx 1.23 \text{ N}\)

The direction \(\theta\) of the resultant force is given by:

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

\(\theta = \arctan\left(\frac{0.562}{-1.098}\right)\)

\(\theta \approx 152.9^\circ \text{ anticlockwise from the positive x-axis}\)

(i) Resolve the forces in the \(x\)-direction:

\(X = 25 \times 0.96 - 30 \times 0.8 = 0\)

Thus, the component in the \(x\)-direction is zero.

(ii) Resolve the forces in the \(y\)-direction:

\(Y = 25 \times 0.28 - 20 + 30 \times 0.6 = 5\)

The resultant has a magnitude of 5 N and acts in the positive \(y\) direction.

(iii) To reverse the direction of the resultant force, the replacement force must have a magnitude of 30 N and act in the negative \(y\) direction.

Resolve the forces into horizontal ( X ) and vertical ( Y ) components:

Horizontal component ( X ):

\(X = 20 \, \cos 60° - 14 - 16 \, \cos 50° = -14.2846 \, \text{N}\)

Vertical component ( Y ):

\(Y = 60 - 20 \, \sin 60° - 16 \, \sin 50° = 30.42278 \, \text{N}\)

Calculate the magnitude of the resultant force ( R ):

\(R = \sqrt{(-14.2846)^2 + (30.42278)^2} = 33.6 \, \text{N}\)

Calculate the direction of the resultant force:

\(\theta = \tan^{-1}\left(\frac{30.42278}{14.2846}\right) = 64.8°\)

The direction is 64.8° above the 14 N force.

(i) To find the y-components of the forces, use the Pythagorean theorem:

\(Y_1^2 = 68^2 - (-60)^2\)

\(Y_1 = \sqrt{68^2 - 60^2} = 32\)

\(Y_2 = 75\) (since the force is vertical)

\(Y_3^2 = 100^2 - 96^2\)

\(Y_3 = \sqrt{100^2 - 96^2} = 28\)

Thus, the components in the positive y-direction are -32, 75, and -28.

(ii) The resultant force components are:

\(X = -60 + 0 + 96 = 36\)

\(Y = -32 + 75 - 28 = 15\)

The magnitude of the resultant force is:

\(R = \sqrt{X^2 + Y^2} = \sqrt{36^2 + 15^2} = 39 \text{ N}\)

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

\(\theta = \arctan \left( \frac{Y}{X} \right) = \arctan \left( \frac{15}{36} \right) \approx 22.6^\circ\)

The direction is 22.6° anticlockwise from the positive x-axis.

Resolve the forces into their components along the x and y axes.

For the x-direction:

\(X = 12\cos 25^{\circ} - 8\cos 10^{\circ} = 2.9972\)

For the y-direction:

\(Y = 12\sin 25^{\circ} + 8\sin 10^{\circ} - 2 = 4.4606\)

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

\(R = \sqrt{X^2 + Y^2} = \sqrt{(2.9972)^2 + (4.4606)^2} = 5.37\)

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

\(\tan \theta = \frac{X}{Y} = \frac{2.9972}{4.4606}\)

\(\theta = \arctan\left(\frac{2.9972}{4.4606}\right) = 33.9^{\circ}\)