Mechanics: Statics and Dynamics

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These are the solutions to the WASSCE Further Mathematics past questions on the topics: Mechanics.
This consists of topics in Statics and Dynamics.
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Symbols and Formulas

(1.) Basic Formulas

$ W = mg \\[3ex] where \\[3ex] W = weight\;(N) \\[3ex] g = \text{acceleration due to gravity}\; (ms^{-2}) \\[3ex] m = \text{mass}\;(kg) \\[3ex] $ (2.) Basic Equations of Motion

$ (a.)\;\; v = u + at \\[3ex] (b.)\;\; s = ut + \dfrac{1}{2}at^2 \\[5ex] (c.)\;\; v^2 = u^2 + 2as \\[5ex] \underline{\text{Working towards gravity}} \\[3ex] a = g \\[5ex] \underline{\text{Working aganist gravity}} \\[3ex] a = -g \\[5ex] $ where:
u = initial velocity
v = final velocity
a = acceleration
t = time
s = distance
g = acceleration due to gravity

(3.) Total Momentum

$ m_1 = \text{mass of first body} \\[4ex] v_1 = \text{velocity of first body} \\[4ex] m_2 = \text{mass of second body} \\[4ex] v_2 = \text{velocity of second body} \\[5ex] \underline{\text{Before Collision: Moving in Same Direction}} \\[4ex] \text{momentum of first body} = m_1v_1 \\[4ex] \text{momentum of second body} = m_2v_2 \\[4ex] \text{total momentum} = m_1v_1 + m_2v_2 \\[5ex] \underline{\text{After Collision: If Stuck Together}} \\[3ex] \text{mass of: first body and second body} = m_1 + m_2 \\[4ex] \text{common velocity} = v \\[3ex] \text{total momentum} = v(m_1 + m_2) \\[5ex] \underline{\text{Based on the Principle of Conservation of Momentum}} \\[3ex] \text{total momentum before collision = total momentum after collision} \\[3ex] m_1v_1 + m_2v_2 = v(m_1 + m_2) \\[5ex] \underline{\text{Before Collision: Moving in Opposite Directions}} \\[4ex] \text{momentum of first body} = m_1v_1 \\[4ex] \text{momentum of second body} = m_2 * -v_2 = -m_2v_2 \\[4ex] \text{The negative value of the velocity is because the second body is moving in opposite direction} \\[3ex] \text{total momentum} = m_1v_1 - m_2v_2 \\[5ex] \underline{\text{After Collision: If Stuck Together}} \\[3ex] \text{mass of: first body and second body} = m_1 + m_2 \\[4ex] \text{common velocity} = v \\[3ex] \text{total momentum} = v(m_1 + m_2) \\[5ex] \underline{\text{Based on the Principle of Conservation of Momentum}} \\[3ex] \text{total momentum before collision = total momentum after collision} \\[3ex] m_1v_1 - m_2v_2 = v(m_1 + m_2) \\[5ex] $ (4.) Forces acting on a body

$ F_x = F \cos\theta \\[3ex] F_y = F \sin\theta \\[3ex] $ where:
$F$ = magnitude of the force
$F_x$ = horizontal component (x-component) of the force
$F_y$ = vertical component (y-component) of the force
θ = angle the force makes with the positive x-axis (measured counterclockwise).

(1.) A body of mass 40 kg is placed on a rough inclined plane which makes an angle of 30° with the horizontal.
If a force of 420 N is applied upwards parallel to the plane, find the:
(a.) maximum frictional force that will keep the body in equilibrium;
(b.) coefficient of friction.
[Take g = 10$ms^{-2}$]

Let us represent the informaion diagrammatically

Number 1

m = mass of the body = 40 kg
θ = 30°
$F_A$ = applied force = 420 N
$F_F$ = frictional force

W = weight of the body
This acts vertically downward, forming the longest side of the right triangle (hypotenuse).

$W_\perp$ = perpendicular component of the weight also known as the Normal force
This is the part of the weight that presses the body against the surface of the inclined plane.
It is the side of the triangle that is perpendicular to the incline (adjacent side relative to the 30° angle).

$W_{||}$ = parallel component of the weight also known as the Downslope force
This is the part of the weight that acts down the slope of the incline, trying to pull the body downward.
It is the side of the triangle that is aligned with the incline (opposite side relative to the 30° angle).

Frictional force opposes motion and acts upslope to prevent motion.
Hence in equilibrium, the applied force (which tries to pull the body forward) is equal to the frictional force (which opposes the motion) and the downslope force (which tries to pull the body downward).

μ = coefficient of friction
This is the ratio of the frictional force to the normal force.

$ W = mg \\[3ex] = 40(10) \\[3ex] = 400\;N \\[5ex] \underline{\text{Downslope force}} \\[3ex] W_{||} = W\sin\theta ...SOHCAHTOA \\[3ex] = 400\sin 30^\circ \\[3ex] = 400 (0.5) \\[3ex] = 200\;N \\[5ex] (a.) \\[3ex] \underline{\text{In Equilibrium}} \\[3ex] F_A = F_F + W_{||} \\[3ex] F_F = F_A - W_{||} \\[3ex] = 420 - 200 \\[3ex] = 220\;N \\[5ex] (b.) \\[3ex] \underline{\text{Normal force}} \\[3ex] W_\perp = W\cos\theta ...SOHCAHTOA \\[3ex] = 400\cos 30^\circ \\[3ex] = 400 * \dfrac{\sqrt{3}}{2} \\[5ex] = 200\sqrt{3}\;N \\[5ex] \mu = \dfrac{F_F}{W_\perp} \\[5ex] = \dfrac{220}{200\sqrt{3}} \\[5ex] = \dfrac{11}{10\sqrt{3}} \\[5ex] = \dfrac{11}{10\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = \dfrac{11\sqrt{3}}{30} $

(2.) A body of mass 6 kg moves with a velocity of 7 $ms^{-1}$.
It collides with a second body moving in the opposite direction with a velocity of 5 $ms^{-1}$.
After collision, the two bodies move together with a velocity of 4 $ms^{-1}$.
Find the mass of the second body.

Let the:
First body be P: mass of 6 kg and velocity of 7 m/s
Second body be Q: mass of m kg and velocity of −5 m/s
The negative value of the velocity is because the second body is moving in the opposite direction
Momentum = mass × velocity

$ \underline{\text{Before Collision}} \\[3ex] \text{Momentum of Body P} = 6 \times 7 = 42\;kgm/s \\[3ex] \text{Momentum of Body Q} = m \times -5 = -5m\;kgm/s \\[3ex] \text{Total Momentum} = (42 - 5m)\;kgm/s \\[5ex] \underline{\text{After Collision}} \\[3ex] \text{Mass of P and Q after they stick together} = (6 + m)\;kg \\[3ex] \text{The common velocity} = 4\;m/s \\[3ex] \text{Total Momentum} = 4(6 + m)\;kgm/s \\[5ex] \text{Based on the Principle of Conservation of Momentum:} \\[3ex] 42 - 5m = 4(6 + m) \\[3ex] 42 - 5m = 24 + 4m \\[3ex] 42 - 24 = 4m + 5m \\[3ex] 9m = 18 \\[3ex] m = \dfrac{18}{2} \\[5ex] m = 2\;kg $

(3.) A particle of weight 12 N lying on a horizontal ground is acted upon by forces $F_1 = (10N, 090^\circ)$, $F_2 = (16N, 180^\circ)$, $F_3 = (7N, 030^\circ)$ and $F_4 = (12N, 300^\circ)$.

(a.) Express all the forces acting on the particle as column vectors.
(b.) Find, correct to two decimal places, the magnitude of the:
(i.) resultant force;
(ii.) acceleration with which the particle starts to move.
[Take g = 10$ms^{-2}$]

(a.) The forces and the components of the forces are:

Force, $F$ (N)	Horizontal Component, $F_x$ (N)	Vertical Component, $F_y$ (N)	Column Vector = $\begin{bmatrix} F_x \\[3ex] F_y \end{bmatrix}$
$10$	$ 10 \cos 90^\circ \\[3ex] 0 $	$ 10 \sin 90^\circ \\[3ex] 10 $	$\begin{bmatrix} 0 \\[3ex] 10 \end{bmatrix}$
$16$	$ 16 \cos 180^\circ \\[3ex] -16 $	$ 16 \sin 180^\circ \\[3ex] 0 $	$\begin{bmatrix} -16 \\[3ex] 0 \end{bmatrix}$
$7$	$ 7 \cos 30^\circ \\[3ex] 6.062177826 $	$ 7 \sin 30^\circ \\[3ex] 3.5 $	$\begin{bmatrix} 6.062177826 \\[3ex] 3.5 \end{bmatrix}$
$12$	$ 12 \cos 300^\circ \\[3ex] 6 $	$ 12 \sin 300^\circ \\[3ex] -10.39230485 $	$\begin{bmatrix} 6 \\[3ex] -10.39230485 \end{bmatrix}$

$ (b.)(i) \\[3ex] \text{Resultant Force}, F_{resultant} = F_x + F_y \\[3ex] = \begin{bmatrix} 0 \\[3ex] 10 \end{bmatrix} + \begin{bmatrix} -16 \\[3ex] 0 \end{bmatrix} + \begin{bmatrix} 6.062177826 \\[3ex] 3.5 \end{bmatrix} + \begin{bmatrix} 6 \\[3ex] -10.39230485 \end{bmatrix} \\[10ex] = \begin{bmatrix} 0 + -16 + 6.062177826 + 6 \\[3ex] 10 + 0 + 3.5 + -10.39230485 \end{bmatrix} \\[10ex] = \begin{bmatrix} -3.937822174 \\[3ex] 3.10769515 \end{bmatrix} \\[10ex] |F_{resultant}| = \sqrt{F_x^2 + F_y^2} \\[4ex] = \sqrt{(-3.937822174)^2 + 3.10769515^2} \\[4ex] = \sqrt{25.16421262} \\[3ex] = 5.016394384 \\[3ex] \approx 5.02\;N \\[5ex] (ii.) \\[3ex] W = mg \\[3ex] where \\[3ex] W = weight = 12\;N \\[3ex] g = \text{acceleration due to gravity} = 10\;ms^{-2} \\[3ex] m = \text{mass} = ? \\[3ex] m = \dfrac{W}{g} \\[5ex] m = \dfrac{12}{10} \\[5ex] m = 1.2\;kg \\[5ex] F_{resultant} = ma \\[3ex] where \\[3ex] a = \text{accelaration of the particle} \\[3ex] a = \dfrac{F_{resultant}}{m} \\[5ex] a = \dfrac{5.016394384}{1.2} \\[5ex] a = 4.180328654 \\[3ex] a \approx 4.18\;ms^{-2} $

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