Question

15. 'O' is the centre of a square ABCD. F1, F2, F3, and F4, and four forces acting along the sides AB, BC, CD and AD as shown in figure. What should be magnitude of F4, so that the force torque about 'O'
is zero


1) F, +F2 + F3
2) F,+F2 - F3
3) F-F2 - F3
4) (Fi + F₂ +F₃)/2​

Answer 1

All the forces are are at same perpendicular distance from the point O and let this distance be r.

Torque acting due to force  τ=r×F

Torque by F1  is,  τ1  =F1

r in the anti-clockwise direction.

Torque by F2   is,  τ2 =F2

​ r in the anti-clockwise direction.

Torque by F3 is, τ3  =F3

​  

r in the clockwise direction.

Since net torque acting about point O is zero.

∴  τ=F1  r+F2  r−F3

​r=0  

⟹F3 =F1 +F2

​  

thank you

Answer 2

Answer:

answer is option 1 i.e. F4 = F1 +F2+F3

Explanation:

given ; total torque is zero

and we know that clockwise torque is positive. .hence;

-F1-F2-F3+F4 = 0

hence

F4 = F3 +F2+F1