A rope is attached to a block, as shown in Figure 5.64. The rope pulls on the block with a force of$\mathbf{210}\mathbf{N}$, at an angle of$\mathit{\theta }\mathbf{=}\mathbf{23}\mathbf{°}$to the horizontal (this force is equal to the tension in the rope).

(a) What is the$\mathbf{x}$component of the force on the block due to the rope?

(b) What is the $\mathbf{y}$component of the force on the block due to the rope?

The rope pulls on the block with a force of $\mathbf{210}\mathit{N}$, at an angle of $\mathit{\theta }\mathbf{=}\mathbf{23}\mathbf{°}$to the horizontal (this force is equal to the tension in the rope).

The components of a force represent the combined vertical and horizontal forces that combine to make the resultant force.The explanation of forces act on the block is shown

We are given the force acted by the rope $F_T=210N$with an angle $\theta =23°$.

We will apply the Momentum Principle to predict the motion of systems that are affected by a known net force. The momentum of a system in uniform motion is constant so it does not change with time and the derivative of the momentum to the time equals zero

$d\overrightarrow{p}/dt=0$

The form of the momentum principle, the next force exerted on the object must equal the change in the momentum and as the change in the momentum is zero, therefore, the net force exerted on the object is zero

${\overrightarrow{F}}_{net}=d\overrightarrow{p}/dt=0$

This result of the net force will help us to get the force on the block in each direction. In -$x$component, we have one force $F_x$, so the change in the momentum in this component will equal $F_x$. The angle between ${\overrightarrow{F}}_T$and ${\overrightarrow{F}}_x$is $\theta$, so the force in $x$-component equals

$F_{x}d{\overrightarrow{p}}_x/dt={\overrightarrow{F}}_T\cos \theta$

…… (1)

Apply equation $\left(1\right)$for$y$-component. We have one force in the$y$-direction$F_y$. Also, the angle between ${\overrightarrow{F}}_N$and $F_y$is $\left(90°-\theta \right)$, so equation for -component will be in the form

$$\begin{gathered} F_y=d{\overrightarrow{p}}_y/dt \\ =\overrightarrow{F}\cos \left(90°-\theta \right) \end{gathered}$$ …… (2)

Now we can plug our values for ${\overrightarrow{F}}_N$and $\theta$into equation $\left(2\right)$to get the $y$-component of the force,

$$\begin{gathered} F_{\mathrm{y}}={\overrightarrow{\mathrm{F}}}_{\mathrm{N}}\cos \left(90°-\mathrm{\theta }\right) \\ =\left(210\mathrm{N}\right)\cos \left(90°-23\right) \\ =82.05\mathrm{N} \end{gathered}$$

Therefore, $y$component of the force on the block due to rope is${\mathrm{F}}_{\mathrm{y}}=82.05\mathrm{N}$.

$F_x=F\cos \theta$

Here,$F_x$is the horizontal or the$x$component of the force acting on the block,$\theta$is the angle,$F$is the tension of the rope.

Substitute $210N$for $F,23°$for $\theta$in equation (1)

$$\begin{gathered} {\mathrm{F}}_{\mathrm{x}}=\left(210\mathrm{N}\right)\cos 23° \\ =193.3\mathrm{N} \end{gathered}$$

Therefore, the $x$component of the force on the block due to the rope is$193.3\mathrm{N}$.