Find the magnitude of force:
Q: The two parallel forces are acting at point A and B respectively is equivalent to force of 100 N acting downwards at point C and couple of 200Nm. Find the magnitude and force F1 and F2 shown in the figure a.
The given system is converts to single force and single couple at point C. Let R be the resultant of forces F1 and F2.
R_{H} = 0
R_{V} = -F_{1} -F_{2}
R_{V} = -(F_{1} + F_{2}) ...(i)
Since R = (R^{2}_{v} )^{1/2} = (F_{1} + F_{2})
Let the resultant R act at distance 'd' from point C.
Now the single force that is R is converted in to single force and couple at C
Now apply two equal and opposite force i.e. 'R' at point C. as shown in the figure b. Now force 'R' which act at point E and upward force which act at point C makes a couple of magnitude = Force × distance
= R × d
But R.d = 200 N-m
And force R which is downward direction = 100 N (negative for downward)
That is, (F_{1} + F_{2}) = 100 or F_{1} + F_{2} = 100 ...(ii)
Now taking moment about C, or apply varignon's theorem.
R.d = 4F_{1} + 7F_{2} , but R.d = 200,
4F_{1} + 7F_{2} = 200 ...(iii)
By solving equation (ii)and (iii)
F_{1} = 500/3 N .......ANS
F_{2 }= -200/3 N .......ANS