Question

A sign with a mass of 10 kg is hung using two cables. The tension in cable A is 85N & the tension in cable B is 97N. Cable A makes an angle of 23 degree with the ceiling. Determine the angle that Cable B makes with the ceiling. Include a diagram to support your solution. Round your answer to 1 decimal place.

Answer 1

We have those physical quantities

m = 10 kg        $T_1 = 85$ N           $T_2 = 97$ N           $\alpha = 23^o$   we find      $\beta$ - ?

$T_1^2 + T_2^2 + 2T_1T_2cos\theta = (mg)^2$

$cos\theta = \large\frac{(mg)^2-T_1^2 - T_2^2}{2T_1T_2} = \large\frac{100^2-97^2-85^2}{2*85*97}=-0.402$

$\theta = 114^o$

$\alpha +\beta+\theta= \pi$

$\beta = \pi - \alpha-\theta = 180^o-23^o-114^o=43^o$

Answer: the angle that Cable B makes with the ceiling is $43^o$