Question

A block is acted on by two forces as shown in the diagram below.If the magnitudes of the forces are F1 = 53.5 N and F2 = 26.0 N, what are the magnitude (in m/s2) and direction of the acceleration of the block? Let m = 8.00 kg and = 41.0°.magnitude m/s2direction

Answer 1

Given data:

* The value of forces acting on the block is,

$\begin{gathered} F_1=53.5\text{ N} \\ F_2=26\text{ N} \end{gathered}$

* The mass of the block is m = 8 kg.

* The angle of the force F_2 with the horizontal is,

$\theta=41^{\circ}$

Solution:

The horizontal component of the force F_2 is,

$\begin{gathered} F_{2x}=-F_2\cos (\theta) \\ F_{2x}=-26\times\cos (41^{\circ}) \\ F_{2x}=-19.62\text{ N} \end{gathered}$

Here, the negative sign indicates the direction of the horizontal force is towards the negative of the x-axis.

The vertical component of the force F_2 is,

$\begin{gathered} F_{2y}=F_2\sin (\theta) \\ F_{2y}=-26\sin (41^{\circ}) \\ F_{2y}=-17.06\text{ N} \end{gathered}$

There is no motion of the block takes place in the vertical direction, thus, the normal force acting on the block is,

$F_N=F_{2y}+mg$

Thus, the net force acting in the vertical direction is,

$\begin{gathered} F_y=F_N-F_{2y}-mg \\ F_y=F_N-(F_{2y}+mg)_{} \\ F_y=0\text{ N} \end{gathered}$

According to Newton's second law, the acceleration of the block in the vertical direction is,

$\begin{gathered} F_y=ma_y \\ a_y=\frac{F_y}{m} \\ a_y=0ms^{-2} \end{gathered}$

The net force acting on the block in the horizontal direction is,

$\begin{gathered} F_x=F_1+F_{2x}_{} \\ F_x=53.5-19.62 \\ F_x=33.88\text{ N} \end{gathered}$

According to newton's second law, the acceleration of the block along the horizontal direction is,

$\begin{gathered} F_x=ma_x \\ a_x=\frac{F_x}{m} \end{gathered}$

Substituting the known values,

$\begin{gathered} a_x=\frac{33.88}{8} \\ a_x=4.235ms^{-2} \end{gathered}$

The magnitude of the acceleration is,

$\begin{gathered} a=\sqrt[]{a^2_x+a^2_y} \\ a=\sqrt[]{(4.235)^2+0} \\ a=4.235ms^{-2} \end{gathered}$

The direction of the acceleration is,

$\begin{gathered} \tan (\theta)=\frac{a_y}{a_x} \\ \tan (\theta)=0 \\ \theta=0^{\circ} \end{gathered}$

Thus, the magnitude of the acceleration is 4.325 meters per second squared and the direction of the acceleration is towards the right (positive x-axis).