In a lever, the effort arm is two times as long as the load arm. The resultant force will be

So we want to know what the resultant force would be if on the lever the effort arm is as twice as long as the load arm. So in order for the lever to be in a state of balance it must be: F1*r1=F2*r2 where F1 is the effort arm and F2 is the load arm, r1 is the perpendicular distance from the force F1 to the pivot and r2 is the perpendicular distance from the force F2 to the pivot. For both sides of the equation to be equal, since r1=2*r2, F1*2*r2=F2*r2. Now we can divide by r2 and we get: 2F1=F2. When we divide by 2: F1 = (1/2) * F2. So the force of the hand that is trying to lift the load is twice as small as the force of the load we are trying to lift.