Question

solve this ....☺☺☺

give proper answer

Answer 1

Solution:  
23 :  Component of $F_{1}$ along x-axis  is $F_{1}(cos(30\degree)$ .
     Component of $F_{2}$ along x-axis is $F_{2} cos(30\degree)$ .
Here , $F_{2}$ is in negative x-direction .
So, 
[tex] F_{x} = F_{1} cos(30\degree) - F_{2} cos(30\degree) [/tex]
                 = $20 cos(30\degree) -10cos(30\degree)$
                  = $20* \frac{ \sqrt{3} }{2} - 10* \frac{ \sqrt{3} }{2} \\ = 10 \sqrt{3} -5 \sqrt{3} \\ =5 \sqrt{3}$ N


24 :  Component $F_{1}$ and $F_{2}$ along y-axis are 
$20 sin(30\degree)$ and $10 sin(30\degree)$ respectively .

Hence, 
$F_{y} = 20 sin(30\degree) + 10 sin(30\degree) \\ F_{y} = 20 * \frac{1}{2} + 10* \frac{1}{2} = 10+5 = 15N$

25 :  The magnitude of resultant of $F_{1}$ and $F_{2}$
    :

$F_{net} = \sqrt{ F_{x}^{2}+ F_{y}^{2} } \\ =\ \textgreater \ \sqrt{ (5 \sqrt{3} )^{2}+ (15)^{2} } \\ =\ \textgreater \ \sqrt{75+225} = \sqrt{300} = 10 \sqrt{3}$N


26: By Looking at the pic, we observe that 
    
$tan(\theta) = \frac{F_{y}}{F_{x}} = \frac{5 \sqrt{3} }{15} = \sqrt{3}$

Hence, 
$\theta = 60\degree$