# Answer pllzz ♥️♥️
Answers
Answer: The resultant torque is 110 N-m.
Explanation:
Given that,
Force F_{A}= 20 NF
A
=20N
Force F_{B}= 30 NF
B
=30N
Force F_{C}= 80 NF
C
=80N
Force F_{D}= 200 NF
D
=200N
Force F_{E}= 50 NF
E
=50N
Distance OB = 4 m
Distance OC = 3 m
Distance OA = 1 m
Distance OE = 5 m
We know that,
The torque is the product of the force and distance.
According to figure,
The torque at D is zero because this force passing from center of mass.
Force B, and force E are clockwise.
So, The torque is
\tau_{B} = 30\ N\times4mτ
B
=30 N×4m
\tau_{B} = 120\ N-mτ
B
=120 N−m
The torque is
\tau_{E} = 50\ N\times5mτ
E
=50 N×5m
\tau_{E} = 250\ N-mτ
E
=250 N−m
Force A and force C are anticlockwise.
So, The torque is
\tau_{A} = 20\ N\times1mτ
A
=20 N×1m
\tau_{A} = 20\ N-mτ
A
=20 N−m
The torque is
\tau_{C} = 80\ N\times3mτ
C
=80 N×3m
\tau_{C} = 240\ N-mτ
C
=240 N−m
Now, the resultant torque is
Resultant torque = clockwise torque - anticlockwise torque
\tau = \tau_{B}+\tau_{E}-\tau_{A}+\tau_{C}τ=τ
B
+τ
E
−τ
A
+τ
C
\tau=120+250-240+20τ=120+250−240+20
\tau= 110\ N-mτ=110 N−m
okkk .....
Hence, The resultant torque is 110 N-m.
as we know,torque is the force to the perpendicular distance from the line of force
further see the attachment....
hope it helps
