Question

A body of mass $10kg$ is acted upon by two perpendicular forces, $6N$ and $8N$. The resultant acceleration of the body is

• A. $1m/s^2$ at an angle of $tan-1\left(\frac{4}{3}\right)$ w.r.t. $6N$ force.
• B. $1m/s^2$ at an angle of $tan-1\left(\frac{3}{4}\right)$ w.r.t. $8N$ force.
  
• C. $0.2m/s^2$ at an angle of $tan-1\left(\frac{3}{4}\right)$ w.r.t. $8N$ force.
• D. $0.2m/s^2$ at an angle of $tan-1\left(\frac{4}{3}\right)$ w.r.t. $6N$ force.
  

Answer 1

Answer: (A), (B)

$1m/s^2$ at an angle of $tan-1\left(\frac{3}{4}\right)$ w.r.t. $8N$ force.


Step 1: Find magnitude of the resultant force.

Formula used: $F=\sqrt{(F_1{)}^2+(F_2{)}^2+2F_1{F}_{2}cos\theta }$

Given, force $F_1=6N$ and force $F_2=8N$

As these two forces are perpendicular to each other,

Angle between the forces,$\theta ={90}^{\circ }$

Resultant force $F=\sqrt{(F_1{)}^2+(F_2{)}^2+2F_1{F}_{2}cos\theta }$

$F=\sqrt{(6{)}^2+(8{)}^2+2\left(6\right)\left(8\right)cos\left({90}^{\circ }\right)}$

$F=\sqrt{100}=10N$

Step 2: Find magnitude of acceleration.

Formula used: $a=\frac{F}{m}$

Given, mass of the body,$m=10kg$

Acceleration of the body, $a=\frac{F}{m}$

$a=\frac{10}{10}=1m/s^2$

Step 3: Find direction of acceleration.

Formula used: $tan\theta =\frac{F_y}{F_x}$

As we know, acceleration lies in the direction of force.

Let acceleration makes $\theta 1$ angle with $F_1\left(F_x\right)$

$tan{\theta }_{}1=\frac{8}{6}=\frac{4}{3}\theta 1=ta{n}^{-1}\left(\frac{4}{3}\right)$

Let acceleration makes $\theta 2$ angle with $F_2\left(F_y\right)$

$tan{\theta }_{}2=\frac{6}{8}=\frac{3}{4}{\theta }_1=ta{n}^{-1}\left(\frac{3}{4}\right)$

Final answer: (a), (c)