Question #9109a

Question

5243 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T12:43:11

$T_1=541.16N$

$T_2~~710.35N$

$T_3=414.55N$

enter image source here
From the figure
$"Upward vertical component of " T_1 =>T_1sin40$

$"Horizontal component of " T_1 =>T_1cos40$

$"Wt of the plank at the mid point"=34xx9.8N$

$"Wt of the man at the point 0.5m from left end"=725N$

$"Leftward horizontal force"=T_3$

$"Upward vertical force on rope at left end" =T_3$

The system is in equlibrium

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

$T_3=T_1cos40...(1)$

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

$T_2+T_1sin40=(725+34xx9.8)N$

$=>T_2+T_1sin40=1058.2N....(2)$

Cosidering the moments of forces about left end we get

$2xxT_1sin40=0.5xx725+1xx(34xx9.8)$

$=>T_1=695.7/(2sin40)N=541.16N$

Inserting the value of $T_1$ in (1)

$T_3=541.16xxcos40=414.55N$

Inserting the value of $T_1$ in (2)

$=>T_2+T_1sin40=1058.2N$

$=>T_2+541.16xxsin40=1058.2N$

$=>T_2=1058.2-541.16xxsin40~~710.35N$