# Mechanics particle and tension help please?

## A particle P of weight 20 N is suspended by two strings AP and BP. The tension in string AP is (ai + 8j ) and the tension in string BP is (6i+bj). The value of a is _ and the value of b is __ The magnitude of the tension in string AP is __ The angle that string BP makes with the vertical is _ Give your answers as exact whole numbers or as decimals to 3 s.f.

Mar 30, 2018

See below.

#### Explanation:

Calling

${F}_{1} = \left(a , 8\right)$
${F}_{2} = \left(6 , b\right)$
and
$P = \left(0 , - m g\right)$

we have after equilibrium

${F}_{1} + {F}_{2} + P = 0$ or

$\left\{\begin{matrix}a + 6 = 0 \\ 8 + b - m g = 0\end{matrix}\right.$

solving for $a , b$ gives

$a = - 6$
$b = m g - 8$

now

${T}_{A P} = \sqrt{{a}^{2} + {8}^{2}} = \sqrt{{6}^{2} + {8}^{2}}$
${T}_{B P} = \sqrt{{6}^{2} + {b}^{2}} = \sqrt{{6}^{2} + {\left(m g - 8\right)}^{2}}$

the anfle between ${F}_{1} , {F}_{2}$ is obtaining by knowing

$\left\langle{F}_{1} , {F}_{2}\right\rangle = \left\lVert {F}_{1} \right\rVert \left\lVert {F}_{2} \right\rVert \cos \phi$ then

$\phi = \arccos \left(\frac{\left\langle{F}_{1} , {F}_{2}\right\rangle}{\left\lVert {F}_{1} \right\rVert \left\lVert {F}_{2} \right\rVert}\right) = \arccos \left(\frac{a \times 6 + 8 \times b}{\sqrt{{a}^{2} + {8}^{2}} \sqrt{{6}^{2} + {b}^{2}}}\right)$