# Question 31f43

Jan 16, 2018

$6.874 \frac{m}{s} ^ 2$ about 71.90 degrees w.r,t north

#### Explanation:

$15 N$ force is acting along south and $50 N$ along northeast. So total force acting along north is $50 \cos 45$ or,35.35 N,so net force acting along north is $\left(35.35 - 14\right)$ or,$21.35 N$

And force acting along east is $\left(50 \cos 45 + 30\right)$ or,$65.35 N$.

so,let ,net force acting along $\theta$ degrees w.r,t North and its magnitude is sqrt#$\left({21.35}^{2} + {65.35}^{2}\right) N$ or,$68.74 N$

$\tan \theta = \frac{65.35}{21.35}$ or,$\theta = 71.90$ degrees

so,magnitude of acceleration will be $\frac{68.74}{10}$ or $6.874 \frac{m}{s} ^ 2$ along the direction of net force.