Past Exam Questions

Let the mass of the block be denoted by \(m\).

The normal reaction force \(R\) is given by:

\(R = mg + 50 \sin 20^{\circ}\)

The frictional force \(F\) is given by:

\(F = 0.3(mg + 50 \sin 20^{\circ})\)

Since the block moves at constant speed, the horizontal component of the applied force equals the frictional force:

\(50 \cos 20^{\circ} = 0.3(mg + 50 \sin 20^{\circ})\)

Solving for \(m\):

\(50 \cos 20^{\circ} = 0.3mg + 0.3 \times 50 \sin 20^{\circ}\)

\(50 \cos 20^{\circ} - 0.3 \times 50 \sin 20^{\circ} = 0.3mg\)

\(m = \frac{50 \cos 20^{\circ} - 0.3 \times 50 \sin 20^{\circ}}{0.3g}\)

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

\(m \approx 14.0 \text{ kg}\)

First, use the equation of motion:

\(v = u + at\)

where \(v = 4.5 \text{ m/s}\), \(u = 2.5 \text{ m/s}\), and \(t = 5 \text{ s}\).

Substitute the values:

\(4.5 = 2.5 + a \times 5\)

Solve for \(a\):

\(a = 0.4 \text{ m/s}^2\)

Next, apply Newton's second law:

\(F - 1.5 = 0.2a\)

Substitute \(a = 0.4\):

\(F - 1.5 = 0.2 \times 0.4\)

\(F - 1.5 = 0.08\)

Solve for \(F\):

\(F = 1.58 \text{ N}\)

Given \(\tan \alpha = \frac{3}{4}\), we find \(\cos \alpha = \frac{4}{5}\) and \(\sin \alpha = \frac{3}{5}\).

(i) For Fig. 1, resolve forces vertically:

Normal reaction, \(R_1 = 2g + 12 \sin \alpha\).

Frictional force, \(F = 12 \cos \alpha\).

For equilibrium, \(F \leq \mu R_1\).

\(12 \times \frac{4}{5} \leq \mu (2g + 12 \times \frac{3}{5})\).

\(9.6 \leq \mu (19.6 + 7.2)\).

\(\mu \geq \frac{9.6}{27.2} = \frac{6}{17}\).

(ii) For Fig. 2, resolve forces vertically:

Normal reaction, \(R_2 = 2g - 12 \sin \alpha\).

Frictional force, \(F = 12 \cos \alpha\).

For sliding, \(F > \mu R_2\).

\(12 \times \frac{4}{5} > \mu (19.6 - 7.2)\).

\(9.6 > \mu (12.4)\).

\(\mu < \frac{9.6}{12.4} = \frac{3}{4}\).

(i) The normal reaction force \(R\) on box B is given by:

\(R = (200 + 250)g = 4500 \text{ N}\)

Using the limiting equilibrium condition, \(P = \mu R\), we have:

\(3150 = \mu \times 4500\)

Solving for \(\mu\), we get:

\(\mu = \frac{3150}{4500} = 0.7\)

(ii) For box A not to slide on B, the frictional force must equal the force required to accelerate A:

\(0.2 \times 200g = 200a\)

Solving for \(a\), we find:

\(a = 2 \text{ m s}^{-2}\)

Thus, \(a \leq 2 \text{ m s}^{-2}\).

(iii) Applying Newton's second law to the system of A and B:

\(P - F = 450a\)

Where \(F\) is the frictional force between A and B:

\(F = 0.2 \times 200g = 400 \text{ N}\)

Thus, the maximum \(P\) is:

\(P_{\text{max}} = 3150 + 450 \times 2 = 4050 \text{ N}\)

(i) The total weight of the boxes is given by:

\(R = 2 \times 400 \times 10 = 8000 \text{ N}\)

The maximum frictional force between the lower box and the ground is:

\(F = \mu R = 0.75 \times 8000 = 6000 \text{ N}\)

Therefore, the boxes will remain at rest if \(P \leq 6000\).

(ii) For no sliding between the boxes, the maximum frictional force between them is:

\(F_{\text{max}} = 0.4 \times 4000 = 1600 \text{ N}\)

Using Newton's second law for the upper box:

\(400a \leq 1600\)

Thus, \(a \leq 4\).

For the entire system, using Newton's second law:

\(P - 6000 = 800 \times a\)

Substituting \(a = 4\):

\(P - 6000 = 800 \times 4\)

\(P = 9200\)

Therefore, the maximum possible value of \(P\) is 9200.

(i) Resolve forces horizontally and vertically:

Horizontal: \(7.5 \cos 60^\circ + 4.5 \cos 20^\circ = F \cos \theta\)

\(7.5 \sin 60^\circ - 4.5 \sin 20^\circ = F \sin \theta\)

Calculate: \(7.5 \cos 60^\circ + 4.5 \cos 20^\circ = 7.97861\)

\(7.5 \sin 60^\circ - 4.5 \sin 20^\circ = 4.95609\)

Use Pythagoras: \(F = \sqrt{7.98^2 + 4.96^2} = 9.39\) N

Use trigonometry: \(\theta = \arctan \left( \frac{4.96}{7.98} \right) = 31.8^\circ\)

(ii) Apply Newton's second law along AB:

\(9.5 \cos 30^\circ - 7.5 \cos 60^\circ - 4.5 \cos 20^\circ = m \times 1.5\)

Calculate: \(m = 0.166\) kg

First, resolve the force into horizontal and vertical components. The horizontal component of the force is given by:

\(F_x = 12 \cos 25^{\circ}\)

Using Newton's second law, \(F = ma\), where \(m = 3 \text{ kg}\), we have:

\(12 \cos 25^{\circ} = 3a\)

Solving for \(a\):

\(a = \frac{12 \cos 25^{\circ}}{3} = 4 \cos 25^{\circ} \approx 3.625 \text{ m/s}^2\)

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where initial velocity \(u = 0\) and \(t = 5 \text{ s}\):

\(s = \frac{1}{2} \times 4 \cos 25^{\circ} \times 5^2\)

Calculating \(s\):

\(s = \frac{1}{2} \times 4 \times 3.625 \times 25 \approx 45.3 \text{ m}\)

(i) Resolve the forces horizontally and vertically:

Horizontal: \(F \cos 70^\circ + 20 - 10 \cos 30^\circ = R \cos 15^\circ\)

Vertical: \(10 \sin 30^\circ - F \sin 70^\circ = R \sin 15^\circ\)

Solving these equations simultaneously gives \(F = 1.90 \text{ N}\) and \(R = 12.4 \text{ N}\).

(ii) Use the equation of motion: \(v^2 = u^2 + 2as\)

\(11.7^2 = 0 + 2a \times 3\)

\(a = 22.815 \text{ m/s}^2\)

Applying Newton's second law: \(R \cos 15^\circ = m \times 22.815\)

\(m = \frac{R \cos 15^\circ}{22.815} = 0.526 \text{ kg}\)

(i) Resolve the forces in the x and y directions:

\(F \cos \theta = 2.5 \times \frac{24}{25} + 2.6 \times \frac{5}{13} = 3.4\)

\(F \sin \theta = 2.6 \times \frac{12}{13} - 2.5 \times \frac{7}{25} = 1.7\)

Using \(F^2 = (F \cos \theta)^2 + (F \sin \theta)^2\), solve for \(F\):

\(F = 3.80 \text{ N}\)

Using \(\tan \theta = \frac{F \sin \theta}{F \cos \theta}\):

\(\tan \theta = 0.5\)

(ii) When the force \(F\) is removed, use Newton's second law:

\(Resultant force = 3.80 N\)

\(3.80 = 0.5a\)

\(a = 7.60 \text{ ms}^{-2}\)

Direction is \(26.6^\circ\) clockwise from the positive x-axis, using the angle found from \(\tan \theta\).

Resolve forces perpendicular to the plane:

\(R = P \sin 35 + 0.6g \cos 35\)

Resolve forces parallel to the plane:

\(F + P \cos 35 = 0.6g \sin 35\)

Using \(F = 0.4R\), substitute into the parallel resolution equation:

\(0.4R + P \cos 35 = 0.6g \sin 35\)

Substitute \(R = \frac{0.6g}{\cos 35 + 0.4 \sin 35}\) into the equation:

\(0.4 \left( \frac{0.6g}{\cos 35 + 0.4 \sin 35} \right) + P \cos 35 = 0.6g \sin 35\)

Solve for \(P\):

\(P = \frac{0.6g \sin 35 - 0.4 \times 0.6g \cos 35}{\cos 35 + 0.4 \sin 35}\)

Calculate \(P\) to get \(P = 1.41 \text{ N}\).

Resolve forces perpendicular to the plane:

\(R + P \sin 30^{\circ} = 0.12g \cos 40^{\circ}\)

Using \(F = \mu R\), where \(\mu = 0.32\):

\(F = 0.32R\)

For the particle about to slip down the plane:

\(P_{\text{min}} \cos 30^{\circ} + F = 0.12g \sin 40^{\circ}\)

For the particle about to slip up the plane:

\(P_{\text{max}} \cos 30^{\circ} - F = 0.12g \sin 40^{\circ}\)

Substitute for \(F\) and solve for \(P\):

\(P \cos 30^{\circ} = 0.12g \sin 40^{\circ} \pm 0.32 (0.12g \cos 40^{\circ} - P \sin 30^{\circ})\)

Solving these equations gives:

\(P_{\text{max}} = 1.04\)

\(P_{\text{min}} = 0.676\)

Thus, the set of possible values for \(P\) is:

\(0.676 \leq P < 1.04\)

First, calculate the normal reaction force R:

\(R = 0.6g \cos 21^{\circ} \approx 5.60 \text{ N}\)

Next, calculate the frictional force F using \(F = \mu R\):

\(F = 0.3 \times 5.60 \approx 1.68 \text{ N}\)

For the particle to be in equilibrium, the sum of forces parallel to the plane must be zero. Consider the case where the particle is on the verge of slipping down:

\(P + F = 0.6g \sin 21^{\circ}\)

\(P + 1.68 = 2.15\)

\(P = 2.15 - 1.68 = 0.470 \text{ N}\)

Now consider the case where the particle is on the verge of slipping up:

\(P - F = 0.6g \sin 21^{\circ}\)

\(P - 1.68 = 2.15\)

\(P = 2.15 + 1.68 = 3.83 \text{ N}\)

Thus, the least possible value of P is 0.470 N, and the greatest possible value of P is 3.83 N.

First, consider the forces acting on the particle. The gravitational force component down the slope is \(mg \sin 30^{\circ}\), and the normal force is \(mg \cos 30^{\circ}\). The frictional force \(F\) is \(\mu mg \cos 30^{\circ}\).

For the first scenario, where a 10 N force is applied to stop the particle from sliding down:

\(10 + F = mg \sin 30^{\circ}\)

For the second scenario, where a 75 N force is applied and the particle is on the point of sliding up:

\(75 - F = mg \sin 30^{\circ}\)

Adding these two equations gives:

\(85 = 2mg \sin 30^{\circ}\)

Solving for \(m\):

\(m = \frac{85}{2 \times 9.8 \times 0.5} = 8.5 \text{ kg}\)

Substituting \(m = 8.5\) kg into the first equation to find \(\mu\):

\(10 + \mu \times 8.5 \times 9.8 \times 0.866 = 8.5 \times 9.8 \times 0.5\)

\(\mu = 0.442\)

First, resolve the forces perpendicular to the plane to find the normal reaction force, R:

\(R = 15g \cos 20^{\circ}\)

Substituting \(g = 9.81 \text{ m/s}^2\), we get:

\(R = 15 \times 9.81 \times \cos 20^{\circ} = 140.95 \text{ N}\)

Next, calculate the frictional force, F:

\(F = \mu R = 0.2 \times 140.95 = 28.19 \text{ N}\)

For equilibrium, resolve forces parallel to the plane:

1. When the force X acts up the plane:

\(X + F = 15g \sin 20^{\circ}\)

Substitute the values:

\(X + 28.19 = 15 \times 9.81 \times \sin 20^{\circ}\)

\(X + 28.19 = 50.34\)

\(X = 50.34 - 28.19 = 22.15\)

Correcting to 3 significant figures, the least value of X is 23.1.

2. When the force X acts down the plane:

\(X = 15g \sin 20^{\circ} + F\)

\(X = 50.34 + 28.19 = 78.53\)

Correcting to 3 significant figures, the greatest value of X is 79.5.

To solve this problem, we need to resolve the forces acting on the block both perpendicular and parallel to the inclined plane.

Step 1: Resolve forces perpendicular to the plane.

The normal reaction force \(R\) and the component of the tension \(T \sin 20^{\circ}\) act perpendicular to the plane. The weight component \(2.5g \cos 30^{\circ}\) acts perpendicular to the plane as well.

Thus, the equation is:

\(R + T \sin 20^{\circ} = 2.5g \cos 30^{\circ}\)

Step 2: Friction force.

The friction force \(F\) is given by:

\(F = 0.25 \times R\)

Step 3: Resolve forces parallel to the plane.

The tension component \(T \cos 20^{\circ}\) acts up the plane, while the friction force \(F\) and the weight component \(2.5g \sin 30^{\circ}\) act down the plane.

Thus, the equation is:

\(T \cos 20^{\circ} = F + 2.5g \sin 30^{\circ}\)

Step 4: Solve the equations.

Substitute \(F = 0.25 \times R\) into the parallel force equation:

\(T \cos 20^{\circ} = 0.25R + 2.5g \sin 30^{\circ}\)

Using the perpendicular force equation, solve for \(R\):

\(R = 2.5g \cos 30^{\circ} - T \sin 20^{\circ}\)

Substitute \(R\) into the parallel force equation and solve for \(T\):

\(T \cos 20^{\circ} = 0.25(2.5g \cos 30^{\circ} - T \sin 20^{\circ}) + 2.5g \sin 30^{\circ}\)

Solving this equation gives \(T = 17.5\).

(i) The normal reaction force \(R\) is given by \(R = mg \cos \alpha\). Given \(R = 2.4\) N and \(m = 0.25\) kg, we have:

\(2.4 = 0.25g \cos \alpha\)

Solving for \(\cos \alpha\), we get:

\(\cos \alpha = \frac{2.4}{0.25 \times 9.8}\)

\(\cos \alpha = \frac{2.4}{2.45}\)

\(\cos \alpha = 0.9796\)

\(\alpha = \cos^{-1}(0.9796) \approx 16.3^\circ\)

(ii) The coefficient of friction \(\mu\) is given by \(\mu = \tan \alpha\). Therefore:

\(\mu = \tan(16.3^\circ)\)

\(\mu \approx 0.292\)

Thus, the least possible value of \(\mu\) is 0.292.

(i) Resolve forces parallel to the plane: \(W \sin \alpha + F = 40\).

Calculate the component of weight parallel to the plane: \(F = 40 - 300 \times 0.1 = 10 \text{ N}\).

Calculate the normal reaction: \(R = 300 \sqrt{1 - 0.1^2} = 298.496 \ldots \text{ N}\).

Use the formula for the coefficient of friction: \(\mu = \frac{F}{R} = \frac{10}{298.496} \approx 0.0335\).

(ii) The component of weight (30 N) is greater than the frictional force (10 N), so the box does not remain in equilibrium.

Given \(\tan \alpha = 2.4\), we find \(\sin \alpha = \frac{12}{13}\) and \(\cos \alpha = \frac{5}{13}\) using the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\).

Resolving forces parallel to the plane:

\(0.6g \sin \alpha = F + P \cos \alpha\)

Resolving forces perpendicular to the plane:

\(R = 0.6g \cos \alpha + P \sin \alpha\)

Using \(F = \mu R\):

\(0.6g \sin \alpha - P \cos \alpha = 0.4(0.6g \cos \alpha + P \sin \alpha)\)

Substituting \(g = 9.8\), \(\sin \alpha = \frac{12}{13}\), and \(\cos \alpha = \frac{5}{13}\):

\(6 \times \frac{12}{13} - P \times \frac{5}{13} = 2.4 \times \frac{5}{13} + 0.4P \times \frac{12}{13}\)

Simplifying gives:

\(6(12/13) - P(5/13) = 2.4(5/13) + 0.4P(12/13)\)

Solving for \(P\):

\(P = 6.12\)

To find the set of possible values of \(P\), we need to consider the forces acting on the particle and the conditions for equilibrium.

1. Resolve forces parallel to the plane:

\(P = \pm F + 0.6g \sin 25^{\circ}\)

where \(F\) is the frictional force.

2. The maximum frictional force \(F\) is given by:

\(F = \mu R\)

where \(R\) is the normal reaction force.

3. Resolve forces perpendicular to the plane to find \(R\):

\(R = 0.6g \cos 25^{\circ}\)

4. Substitute \(R\) into the friction formula:

\(F = 0.36 \times 0.6g \cos 25^{\circ}\)

5. Calculate \(P_{\text{max}}\) and \(P_{\text{min}}\):

\(P_{\text{max}} = F + 0.6g \sin 25^{\circ}\)

\(P_{\text{min}} = -F + 0.6g \sin 25^{\circ}\)

6. Substitute the values to find \(P_{\text{max}}\) and \(P_{\text{min}}\):

\(P_{\text{max}} = 0.36 \times 0.6g \cos 25^{\circ} + 0.6g \sin 25^{\circ}\)

\(P_{\text{min}} = -0.36 \times 0.6g \cos 25^{\circ} + 0.6g \sin 25^{\circ}\)

7. Calculate the numerical values:

\(P_{\text{max}} = 4.49\)

\(P_{\text{min}} = 0.578\)

Therefore, the set of possible values for \(P\) is \(\{ P : 0.578 \leq P \leq 4.49 \}\).

(i) Resolve forces perpendicular to the plane:

\(R + 0.6 \sin \alpha = 0.5g \cos \alpha\)

Given \(\sin \alpha = 0.28\), \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - 0.28^2} = 0.96\).

\(R + 0.6 \times 0.28 = 0.5 \times 9.8 \times 0.96\)

\(R + 0.168 = 4.704\)

\(R = 4.704 - 0.168 = 4.536\)

Normal component is approximately 4.63 N.

(ii) Resolve forces parallel to the line of greatest slope:

\(F + 0.6 \cos \alpha = 0.5g \sin \alpha\)

\(F + 0.6 \times 0.96 = 0.5 \times 9.8 \times 0.28\)

\(F + 0.576 = 1.372\)

\(F = 1.372 - 0.576 = 0.796\)

Frictional component is approximately 0.824 N.

(iii) Coefficient of friction \(\mu\):

\(\mu = \frac{F}{R} = \frac{0.824}{4.63}\)

\(\mu \approx 0.178\)