The Workbook

1. \(F_x = F\cos 37^\circ = 10 \times 0.8 = 8\) N.

2. \(F_y = F\sin 37^\circ = 10 \times 0.6 = 6\) N.

3. \(\vec F = 8\hat i + 6\hat j\) N. Check: \(\sqrt{64 + 36} = 10\).

1. \(|\vec A| = \sqrt{3^2 + 4^2} = 5\).

2. \(\tan\theta = \dfrac{4}{3}\), so \(\theta = 53^\circ\).

3. \(\hat A = \dfrac{\vec A}{|\vec A|} = \dfrac{3\hat i + 4\hat j}{5} = 0.6\hat i + 0.8\hat j\).

\(F_x = 20\cos 120^\circ = 20 \times \left(-\dfrac{1}{2}\right) = -10\) N. It points along \(-x\).

\(F_y = 20\sin 120^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3\) N, along \(+y\).

The unit vector along \(3\hat i + 4\hat j\) is \(\dfrac{3\hat i + 4\hat j}{5}\). Multiply by \(10\): \(6\hat i + 8\hat j\).

1. The two legs are perpendicular: \(|\vec s| = \sqrt{6^2 + 8^2} = 10\) km.

2. \(\tan\theta = \dfrac{8}{6} = \dfrac{4}{3}\), so \(\theta = 53^\circ\) north of east.

3. Distance \(= 6 + 8 = 14\) km. Distance adds the lengths; displacement adds the vectors.

1. \(\vec A + \vec B = (2 + 4)\hat i + (3 - 1)\hat j = 6\hat i + 2\hat j\). Magnitude \(= \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \approx 6.3\).

2. \(\vec A - \vec B = (2 - 4)\hat i + (3 + 1)\hat j = -2\hat i + 4\hat j\).

1. Same direction: \(3 + 4 = 7\) N.

2. Opposite directions: \(4 - 3 = 1\) N.

3. Perpendicular: \(\sqrt{9 + 16} = 5\) N.

The resultant of two forces always lies between their difference and their sum.

With \(B = A\): \(R^2 = 2A^2(1 + \cos\theta)\).

1. \(\cos 60^\circ = \dfrac{1}{2}\): \(R^2 = 3A^2\), \(R = \sqrt{3}A\).

2. \(\cos 90^\circ = 0\): \(R = \sqrt{2}A\).

3. \(\cos 120^\circ = -\dfrac{1}{2}\): \(R^2 = A^2\), \(R = A\).

\(R^2 = 36 + 64 + 96\cos\theta = 100 + 96\cos\theta\).

1. \(100 = 100 + 96\cos\theta\) gives \(\cos\theta = 0\): \(\theta = 90^\circ\).

2. \(4 = 100 + 96\cos\theta\) gives \(\cos\theta = -1\): \(\theta = 180^\circ\), opposite directions.

3. \(196 = 100 + 96\cos\theta\) gives \(\cos\theta = 1\): \(\theta = 0^\circ\), same direction.

1. \(R = \sqrt{36 + 36 + 2 \times 36 \times \cos 60^\circ} = \sqrt{108} = 6\sqrt{3} \approx 10.4\) units.

2. The two vectors are equal in magnitude, so the parallelogram is a rhombus. Its diagonal bisects the angle: \(\vec R\) is at \(30^\circ\) to \(\vec A\).

3. \(|\vec A - \vec B|^2 = 36 + 36 - 2 \times 36 \times \cos 60^\circ = 36\), so \(|\vec A - \vec B| = 6\). It is the other diagonal of the rhombus.

1. Along the incline: \(W\sin 30^\circ = 20 \times \dfrac{1}{2} = 10\) N, pointing down the slope.

2. Perpendicular to the incline: \(W\cos 30^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3\) N, pressing into the slope.

3. \(\sqrt{10^2 + (10\sqrt{3})^2} = \sqrt{100 + 300} = 20\) N.

1. \(F_x = 50\cos 37^\circ = 40\) N, forward along the ground.

2. \(F_y = 50\sin 37^\circ = 30\) N. A push along the handle points downward at the roller, so this component presses the roller into the ground.

3. A pull along the handle points upward: the vertical component is \(30\) N upward. It lifts the roller a little, which is why pulling is easier than pushing.

1. \(T_1\) makes \(30^\circ\) with the ceiling on one side and \(T_2\) makes \(60^\circ\) on the other. The angle between them at the knot is \(180^\circ - 30^\circ - 60^\circ = 90^\circ\).

2. The angle between \(T_1\) and the weight is \(90^\circ + 30^\circ = 120^\circ\); between \(T_2\) and the weight it is \(90^\circ + 60^\circ = 150^\circ\).

   Lami: \(\dfrac{T_1}{\sin 150^\circ} = \dfrac{T_2}{\sin 120^\circ} = \dfrac{10}{\sin 90^\circ}\). So \(T_1 = 10 \times \dfrac{1}{2} = 5\) N and \(T_2 = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.7\) N.

3. Horizontal: \(T_1\cos 30^\circ = 5 \times 0.866 = 4.33\) N and \(T_2\cos 60^\circ = 8.66 \times 0.5 = 4.33\) N. They balance. Vertical check: \(5 \times 0.5 + 8.66 \times 0.866 = 2.5 + 7.5 = 10\) N.

1. \(v_x = 20\cos 60^\circ = 10\) m/s; \(v_y = 20\sin 60^\circ = 10\sqrt{3} \approx 17.3\) m/s.

2. \(\vec v = 10\hat i + 10\sqrt{3}\hat j\) m/s.

3. Gravity acts only vertically. So \(v_y\) decreases, becomes zero at the top and then grows downward, while \(v_x\) stays \(10\) m/s throughout.

Lami's theorem: three concurrent forces keep a particle in equilibrium. Then each force is proportional to the sine of the angle between the other two: \(\dfrac{P}{\sin\alpha} = \dfrac{Q}{\sin\beta} = \dfrac{R}{\sin\gamma}\).

For equilibrium the \(5\) N force must balance the resultant of \(3\) N and \(4\) N, so that resultant is \(5\) N. \(25 = 9 + 16 + 24\cos\theta\) gives \(\cos\theta = 0\): the angle between the \(3\) N and \(4\) N forces is \(90^\circ\).

By Lami's theorem the sines of the three angles are equal, and the three angles add to \(360^\circ\). So each angle is \(120^\circ\). Check: two forces \(F\) at \(120^\circ\) have resultant \(F\), which the third force cancels.

1. The third force cancels the sum: \(\vec F_3 = -(6\hat i + 8\hat j) = -6\hat i - 8\hat j\) N.

2. \(|\vec F_3| = \sqrt{36 + 64} = 10\) N.

3. \(\tan\phi = \dfrac{8}{6} = \dfrac{4}{3}\): \(\phi = 53^\circ\) below the \(-x\) direction.

Take east as \(\hat i\) and north as \(\hat j\). The first two forces add to \(10\hat i + 10\hat j\), of magnitude \(10\sqrt{2}\) N towards north-east.

The third force is \(10\sqrt{2}\) N towards south-west, exactly opposite: \(10\sqrt{2}\left(-\dfrac{\hat i + \hat j}{\sqrt{2}}\right) = -10\hat i - 10\hat j\). The three forces add to zero, so the particle is in equilibrium.

\(\vec v_1 = 10\hat i\) (east) and \(\vec v_2 = 10\hat j\) (north). \(\Delta\vec v = 10\hat j - 10\hat i\).

Magnitude \(= \sqrt{100 + 100} = 10\sqrt{2} \approx 14\) m/s. The direction \(-\hat i + \hat j\) points towards north-west.

\(\vec r = 3\hat i + 4\hat j + 12\hat k\) km. \(|\vec r| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\) km.

1. \(|\vec A + \vec B| = \sqrt{25 + 144} = 13\).

2. The resultant and \(\vec B\) form a right triangle with \(\vec A\) as the opposite side: \(\tan\alpha = \dfrac{5}{12}\).

\(|\vec A + \vec B|^2 = A^2 + B^2 + 2AB\cos\theta\) and \(|\vec A - \vec B|^2 = A^2 + B^2 - 2AB\cos\theta\).

Setting them equal gives \(4AB\cos\theta = 0\). Neither magnitude is zero, so \(\cos\theta = 0\) and \(\theta = 90^\circ\). The two diagonals of a rectangle are equal.

Let the common magnitude be \(A\). \(|\vec A + \vec B|^2 = A^2 + A^2 + 2A^2\cos\theta = A^2\).

So \(2 + 2\cos\theta = 1\), \(\cos\theta = -\dfrac{1}{2}\) and \(\theta = 120^\circ\).

First case: \(R^2 = P^2 + Q^2 + 2PQ\cos\theta\). Second case, with \(2Q\): \(R^2 = P^2 + 4Q^2 + 4PQ\cos\theta\).

Subtract the first from the second: \(0 = 3Q^2 + 2PQ\cos\theta\), so \(\cos\theta = -\dfrac{3Q}{2P}\). The angle is obtuse.