Sum moments about any point. I will choose the Right end.
-.3*75g -1*20g+Fleft*2=0
That gives you Fleft end.
Now, sum forces vertically
Fleft+Fright-20g-75g=0
solve for fright.
a person stands a distance of 0.300 meters from the right end of a 2.00 meter long uniform platform that is supported by two posts, one at each end. The board has a mass of 20.0 kg and the person's mass is 75.0 kg.
a. determine the force exterted on the board by the post on the left end.
b. determine the force exerted on the board by the post on the right end.
2 answers
This is a sum of torque problem.
T(left) = F(board)*Distance from left + F(person)*Distance left + F(right)*Distance from left = 0
T(right) = F(board)*dist from right + F(person)*dist from right + F(left)*distance from right.
In reality, we are summing clockwise torques with counter clockwise torques and since there is no motion ( we are in equilibrium) setting this equal to 0.
This also works with the diving board problems, or the stacked books problem (after establishing the center of gravity for the books).
T(left) = F(board)*Distance from left + F(person)*Distance left + F(right)*Distance from left = 0
T(right) = F(board)*dist from right + F(person)*dist from right + F(left)*distance from right.
In reality, we are summing clockwise torques with counter clockwise torques and since there is no motion ( we are in equilibrium) setting this equal to 0.
This also works with the diving board problems, or the stacked books problem (after establishing the center of gravity for the books).
2 answers