# Question #9109a

Nov 21, 2016

${T}_{1} = 541.16 N$

${T}_{2} \approx 710.35 N$

${T}_{3} = 414.55 N$

#### Explanation:

From the figure
$\text{Upward vertical component of } {T}_{1} \implies {T}_{1} \sin 40$

$\text{Horizontal component of } {T}_{1} \implies {T}_{1} \cos 40$

$\text{Wt of the plank at the mid point} = 34 \times 9.8 N$

$\text{Wt of the man at the point 0.5m from left end} = 725 N$

$\text{Leftward horizontal force} = {T}_{3}$

$\text{Upward vertical force on rope at left end} = {T}_{3}$

The system is in equlibrium

So considering the equilbrium of forces in horizontal direction, we can write

${T}_{3} = {T}_{1} \cos 40. . . \left(1\right)$

Considering the equilbrium of forces in vertial direction, we can write

${T}_{2} + {T}_{1} \sin 40 = \left(725 + 34 \times 9.8\right) N$

$\implies {T}_{2} + {T}_{1} \sin 40 = 1058.2 N \ldots . \left(2\right)$

Cosidering the moments of forces about left end we get

$2 \times {T}_{1} \sin 40 = 0.5 \times 725 + 1 \times \left(34 \times 9.8\right)$

$\implies {T}_{1} = \frac{695.7}{2 \sin 40} N = 541.16 N$

Inserting the value of ${T}_{1}$ in (1)

${T}_{3} = 541.16 \times \cos 40 = 414.55 N$

Inserting the value of ${T}_{1}$ in (2)

$\implies {T}_{2} + {T}_{1} \sin 40 = 1058.2 N$

$\implies {T}_{2} + 541.16 \times \sin 40 = 1058.2 N$

$\implies {T}_{2} = 1058.2 - 541.16 \times \sin 40 \approx 710.35 N$