The cg of the 50.0 N pole held by the pole vaulter is 2.00 m from the left hand, and the hands are
0.700 m apart. Calculate the force exerted by each hand.
This is a problem of levers which can be solved by taking moments.
Three forces are involved, L the left hand, R the right hand, and W the weight of the pole equal to 50N.
It is not clear but will be assumed that the cg of the pole is 2.00m to the left of the left hand, as follows.
---------------W---2.0----L--0.7--R----
Take moments about L, then forces W and R are both acting downwards so as not to create a couple.
2.0W=0.7R
Solve for R (knowing that W=50N)
Now take moments about R, i.e. put a fulcrum at R.
To counter the rotation created by the downwards force W, L must act upwards.
2.7W=0.7L
Solve for L.
To calculate the force exerted by each hand, we can use the principle of moments. The principle of moments states that the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.
In this case, let's choose the left hand as the point of calculation. We want to find the forces exerted by each hand, so let's call the force exerted by the left hand F1 and the force exerted by the right hand F2.
First, let's determine the moments about the left hand:
Moment due to the pole = Force × Distance
Moment due to F1 = F1 × 0.700 m
Moment due to F2 = F2 × 1.700 m (2.00 m - 0.700 m)
According to the principle of moments, the sum of clockwise moments is equal to the sum of anticlockwise moments:
0.700 × F1 + 1.700 × F2 = 0 (because the pole is in equilibrium)
Since the pole is not moving vertically, we can also say that the sum of the vertical forces is equal to zero:
F1 + F2 = 50.0 N (the weight of the pole)
Now we have two equations that we can solve simultaneously to find the values of F1 and F2.
Solving the system of equations gives:
F1 = 33.33 N
F2 = 16.67 N
Therefore, the force exerted by the left hand is 33.33 N and the force exerted by the right hand is 16.67 N.