Question

A uniform metre rod of 5 metre length is suspended horizontally by two strings P and Q.Pis attached 0.8 metre from one end and Q is attached 2 metre from the other end.Given that the weight of the rod is 110N. Calculate the tension in each string

Answer 1

Answer:

The tension in string P is 25 N, while that of Q is 85 N.

Explanation:

Considering the conditions for equilibrium,

i. Total upward force = Total downward force

                     $T_{P}$ + $T_{Q}$ = 110 N

ii. Taking moment about P,

clockwise moment = anticlockwise moment

110 × (2.5 - 0.8) = $T_{Q}$ × (3 - 0.8)

110 × 1.7 = $T_{Q}$ × 2.2

187 = 2.2$T_{Q}$

$T_{Q}$ = $\frac{187}{2.2}$

$T_{Q}$ = 85 N

From the first condition,

$T_{P}$ + $T_{Q}$ = 110 N

$T_{P}$ + 85 N = 110 N

$T_{P}$ = 110 - 85

$T_{P}$ 25 N

Therefore, the tension in string P is 25 N while that of Q is 85 N.