Question 1:
Resolve the following forces into horizontal component and vertical component.
Answer:
(a)
$$
\begin{aligned}
& F_x \equiv 70 \cos 42^{\circ} \\
& F_x \equiv 52.02 \mathrm{~N}
\end{aligned}
$$
$$ \begin{aligned} & F_y=70 \sin 42^{\circ} \\ & F_y=46.84 \mathrm{~N} \end{aligned} $$
(b)
$$ \begin{aligned} F_x & =90 \cos 64^{\circ} \\ F_x & \equiv 39.45 \mathrm{~N} \end{aligned} $$
$$ \begin{aligned} & E_y \equiv 90 \sin 64^{\circ} \\ & E_y \equiv 80.89 \mathrm{~N} \end{aligned} $$
Resolve the following forces into horizontal component and vertical component.
Answer:
(a)
$$
\begin{aligned}
& F_x \equiv 70 \cos 42^{\circ} \\
& F_x \equiv 52.02 \mathrm{~N}
\end{aligned}
$$$$ \begin{aligned} & F_y=70 \sin 42^{\circ} \\ & F_y=46.84 \mathrm{~N} \end{aligned} $$
(b)
$$ \begin{aligned} F_x & =90 \cos 64^{\circ} \\ F_x & \equiv 39.45 \mathrm{~N} \end{aligned} $$
$$ \begin{aligned} & E_y \equiv 90 \sin 64^{\circ} \\ & E_y \equiv 80.89 \mathrm{~N} \end{aligned} $$
Question 2:
Figure 1.22 shows a man pushing a lawn mower with a force of 90 N.
(a) Resolve the pushing force into its horizontal component and vertical component.
(b) State the function of the horizontal component and vertical component of the pushing force when the lawn mower is being pushed.
Answer:
(a)
$$ \begin{aligned} &\text { Horizontal component }\\ &\begin{aligned} & F_x=90 \sin 60^{\circ} \\ & F_x=77.94 N \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Vertical component }\\ &\begin{aligned} & F_y=90 \cos 60^{\circ} \\ & F_y=45.00 \mathrm{~N} \end{aligned} \end{aligned} $$
(b)
The horizontal component (Fx) moves the lawn mower forward.
The vertical component (Fy) pushes the lawn mower on the surface of the field.
Figure 1.22 shows a man pushing a lawn mower with a force of 90 N.
(a) Resolve the pushing force into its horizontal component and vertical component.
(b) State the function of the horizontal component and vertical component of the pushing force when the lawn mower is being pushed.
Answer:
(a)
$$ \begin{aligned} &\text { Horizontal component }\\ &\begin{aligned} & F_x=90 \sin 60^{\circ} \\ & F_x=77.94 N \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Vertical component }\\ &\begin{aligned} & F_y=90 \cos 60^{\circ} \\ & F_y=45.00 \mathrm{~N} \end{aligned} \end{aligned} $$
(b)
The horizontal component (Fx) moves the lawn mower forward.
The vertical component (Fy) pushes the lawn mower on the surface of the field.