A uniform 300-n trapdoor in a floor is hinged at one side. find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges. 11.5 .. raising a ladder. a ladder carried

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The force body diagrams are shown in the picture for the two different cases.

In part a, there are 4 forces acting on the system. Two x and y forces in the point of the hinge on the rightside corner. One force pointing downwards for the weight of the trapdoor which is 300 N. One force pointing up which is what is asked. Let's denote this as x. To find this, we have to know the length of the trapdoor. Since it is not given, let's assume it to be 6 meters. Hence, the distance from the hinge to the center is 3 meters. Using the law of conservation of momentum, 

Summation of Moment = 0
Summation of Moment = x(3)  - 300(3) = 0
x = 300 N

In part b, the upward force is at the leftside corner instead of at the center. Using the same procedure,

Summation of Moment = x(6)  - 300(3) = 0
x = 150 N
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