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As shown in figure, the hinges `A` and `B` hold a uniform `400N` door in place. The upper hinge supports the entire weight of the door. Find the resultant force exerted on the door at the hinges, the width of the door is `h/2`, where `h` is the distance between the hinges. A. `312N`B. `280N`C. `412N`D. `480N` |
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Answer» Correct Answer - C Only a horizontal force acts at hinge `B`, because hinge `A` is assumed to support the door, s weight. Let us take torques about `A` as axis. `sumF_(x)=0`, or `F_(2)-H=0` `sum F_(y)=0` or `V-400 N=0` We find from these that `H=100 N` and `V=400 N` To find the resultant force `vec(R)` on hinge at `A`, we have `R=sqrt(((400)^(2)+(100)^(2))=412 N)` |
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