Question #9109a

1 Answer
Nov 21, 2016

Answer:

#T_1=541.16N#

#T_2~~710.35N#

#T_3=414.55N#

Explanation:

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#