Question

A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the

magnitude and direction of the acceleration of the body.
​

Answer 1

Answer:

Explanation:

body of mass m is 5 kg  

R=  

(8)  

2

 

​  

+(−16  

2

)=  

64+36

​  

=10N

θis angle made by force of 8 N

θ=tan  

−1

(−6/8)=−36.87  

0

 

the negative sign indicates theta clockwise direction respect to the force of magnitude 8N

force is m X a

a=f/m=10/5=2ms  

−2

 

Answer 2

Explanation:

${\underline{\underline{\sf{\red{{\bigstar}\:\:\:Given:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\green{A\: body\: of \:mass\: 5 kg\: is \:acted\: upon\: by \:two\: perpendicular\: forces \:8 N\: and\: 6 N }$

${\underline{\underline{\sf{\orange{{\bigstar}\:\:\:ToFind:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\purple{Magnitude \:and \:direction \:of\: the \:acceleration\: of\: the \:body.</p><p>}$

${\underline{\underline{\sf{\blue{{\bigstar}\:\:\: Solution:-}}}}}$

$\tiny\dag\:\sf{The \:resultant\: of\: two\: forces\: is\: given\: as:}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ R= \sqrt{(8)^2 + (6)^2} }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ R= \sqrt{64+36}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{R = 10N}$

$\tiny\dag\:\sf{\theta\:is\: the\: angle\: made\: by\: R \:with \:the \:force\: of\: 8 N}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{\theta= tan^{-1}(\frac{6}{8})= 36.87^{\circ}}$

$\star\:{\underline{\mathfrak{\purple{Using \: 2nd\:law\:of\: motion}}}}:-$

${\red{\underline{\boxed{\sf{F= ma}}}}}$

$\tiny{\underline{\sf{\:\:\:\:Where,\:\:\:}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\orange{F = force}\\\tiny\:\:\:\:\bullet\:\:\:\sf\green{m = mass \ of \ the \ body}\\\tiny\:\:\:\:\bullet\:\:\:\sf\blue{a = acceleration}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{a = \dfrac{F}{m}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{a = \dfrac{10}{5}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{a = 2m/s^2}$