Question 1:
Figure 1 shows the top view of a worker, X who is applying a pulling force of 70 N on a sack of flour on a track. Another worker, Y is able to apply a pulling force of 60 N on the sack. Determine the direction of the pulling force that must be applied by worker Y on the sack so that the sack moves along the line PQ.
[Ignore the friction between the sack and the surface of the track]
Answer:
The resultant force of worker X (70 N) and worker Y (60 N) has to act along the line PQ. The triangle of forces for forces 70 N, 60 N and resultant force, F is as follows:
$$ \begin{aligned} &\text { Using the sine rule, }\\ &\begin{aligned} \frac{70}{\sin \alpha} & =\frac{60}{\sin 40^{\circ}} \\ \sin \alpha & =\frac{70 \times \sin 40^{\circ}}{60} \\ \alpha & =48.58^{\circ} \\ \theta & =180-40-48.58 \\ & =91.42^{\circ} \\ \beta & =180-91.42 \\ & =88.58^{\circ} \end{aligned} \end{aligned} $$
Therefore, worker Y has to apply a force that makes an angle of 88.58° with the direction of the force from worker X.
Figure 1 shows the top view of a worker, X who is applying a pulling force of 70 N on a sack of flour on a track. Another worker, Y is able to apply a pulling force of 60 N on the sack. Determine the direction of the pulling force that must be applied by worker Y on the sack so that the sack moves along the line PQ.
[Ignore the friction between the sack and the surface of the track]
Answer:
The resultant force of worker X (70 N) and worker Y (60 N) has to act along the line PQ. The triangle of forces for forces 70 N, 60 N and resultant force, F is as follows:
$$ \begin{aligned} &\text { Using the sine rule, }\\ &\begin{aligned} \frac{70}{\sin \alpha} & =\frac{60}{\sin 40^{\circ}} \\ \sin \alpha & =\frac{70 \times \sin 40^{\circ}}{60} \\ \alpha & =48.58^{\circ} \\ \theta & =180-40-48.58 \\ & =91.42^{\circ} \\ \beta & =180-91.42 \\ & =88.58^{\circ} \end{aligned} \end{aligned} $$
Therefore, worker Y has to apply a force that makes an angle of 88.58° with the direction of the force from worker X.
Question 2:
Figure 2 shows the top view of a pulling force applied by two persons, P and Q in an attempt to pull down a tree.
(a) By using the method of parallelogram of forces, determine the magnitude and direction of the resultant force on the tree.
(b) Discuss the advantage and disadvantage of having a large angle between the directions of the two forces.
(c) Which person has to be more careful when the tree begins to topple?
Answer:
(a) Use the scale: 1.0 cm = 20 N
Length of the diagonal of the parallelogram = 9.4 cm
Resultant force, F
= 9.4 × 20
= 188 N
F = 188 N at an angle of 33° with the direction of the force applied by P.
(b)
– Advantage: The tree will fall in the direction of the resultant force. A larger angle will ensure that there is a large space between P and Q. The tree will fall on to the ground without endangering P dan Q.
– Disadvantage: The large angle between the directions of the forces produces a resultant force with a smaller magnitude.
(c) The direction of the resultant force makes a smaller angle with the direction of the force by P. The tree will fall nearer to P. Therefore, P has to be more careful.
Figure 2 shows the top view of a pulling force applied by two persons, P and Q in an attempt to pull down a tree.
(a) By using the method of parallelogram of forces, determine the magnitude and direction of the resultant force on the tree.
(b) Discuss the advantage and disadvantage of having a large angle between the directions of the two forces.
(c) Which person has to be more careful when the tree begins to topple?
Answer:
(a) Use the scale: 1.0 cm = 20 N
Length of the diagonal of the parallelogram = 9.4 cm
Resultant force, F
= 9.4 × 20
= 188 N
F = 188 N at an angle of 33° with the direction of the force applied by P.
(b)
– Advantage: The tree will fall in the direction of the resultant force. A larger angle will ensure that there is a large space between P and Q. The tree will fall on to the ground without endangering P dan Q.
– Disadvantage: The large angle between the directions of the forces produces a resultant force with a smaller magnitude.
(c) The direction of the resultant force makes a smaller angle with the direction of the force by P. The tree will fall nearer to P. Therefore, P has to be more careful.
Question 3:
Figure 3 shows a children’s playground equipment. The spring in the equipment experiences a compression of 5.0 cm when a child of mass 28 kg sits on it. What is the spring constant of the spring in N m–1?
[Gravitational acceleration, g = 9.81 m s–2]
Answer:
Force on spring,
$$ \begin{aligned} F & =\text { weight of child } \\ & =m g \\ & =28 \times 9.81 \\ & =274.68 \mathrm{~N} \end{aligned} $$
$$ \text { Compression, } \begin{aligned} x & =5.0 \mathrm{~cm} \\ & =0.05 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} F & =k x \\ k & =\frac{F}{x} \\ & =\frac{274.68}{0.05} \\ & =5493.6 \mathrm{~N} \mathrm{~m}^{-1} \end{aligned} $$
Figure 3 shows a children’s playground equipment. The spring in the equipment experiences a compression of 5.0 cm when a child of mass 28 kg sits on it. What is the spring constant of the spring in N m–1?
[Gravitational acceleration, g = 9.81 m s–2]
Answer:
Force on spring,
$$ \begin{aligned} F & =\text { weight of child } \\ & =m g \\ & =28 \times 9.81 \\ & =274.68 \mathrm{~N} \end{aligned} $$
$$ \text { Compression, } \begin{aligned} x & =5.0 \mathrm{~cm} \\ & =0.05 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} F & =k x \\ k & =\frac{F}{x} \\ & =\frac{274.68}{0.05} \\ & =5493.6 \mathrm{~N} \mathrm{~m}^{-1} \end{aligned} $$