Question

Two forces are acting on a point at an angle of $\alpha$ and $\beta$ as shown in figure. If the sum of these forces is 700 N directed vertically, What are the angles $\alpha$ and $\beta$?

• A. $\alpha ={55.95}^{\circ },\beta ={45.58}^{\circ }$
• B. $\alpha ={45.58}^{\circ },\beta ={55.95}^{\circ }$
• C. $\alpha ={90}^{\circ },\beta ={30}^{\circ }$
• D. $\alpha ={40.166}^{\circ },\beta ={53.73}^{\circ }$

Answer 1

The correct option is A $\alpha ={55.95}^{\circ },\beta ={45.58}^{\circ }$

Net force is vertical,
So, $400sin\beta +500sin\alpha =700$

$4sin\beta +5sin\alpha =7$

Net force is horizontal direction, is zero
$400cos\beta =500cos\alpha$

$cos\beta =1.25cos\alpha$

$sin\beta =\sqrt{1-(1.25cos\alpha {)}^2}$

By equation (i) and (ii)

$4sin\beta =7-5sin\alpha$

$4\sqrt{1-1.5625co{s}^2\alpha }=(7-5sin\alpha {)}^2$

Squaring on both side.

$16(1-1.5625co{s}^2\alpha )=(7-5sin\alpha {)}^2$

$16-25co{s}^2\alpha =49+25si{n}^2\alpha -70sin\alpha$

$49+25-16=70sin\alpha$

$sin\alpha =\frac{58}{70}=0.82857$

$\alpha ={55.95226}^{\circ }$

$cos\beta =\left(1.25\right)cos\alpha$

$=1.25cos\left({55.95226}^{\circ }\right)$

$\beta ={45.5847}^{\circ }$