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A vaulter is holding a horizontal 3.00-kg pole, 4.50 m long. His front arm lifts straight up on the pole, 0.750 from the end, and his back arm pushes straight down on the end of the pole. How much force does his back arm exert on the pole?

(Unit= N)

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58.8 N

Explanation:

Let 'F₁' be the force by front arm and 'F₂' be the force by back arm.

Given:

Mass of the rod (m) = 3.00 kg

Length of the pole (L) = 4.50 m

Acceleration due to gravity (g) = 9.8 m/s²

Distance of 'F₁' from one end of pole (d₁) = 0.750 m

'F₂' acts on the end. So, distance between 'F₁' and 'F₂' = 0.750 m

Now, weight of the pole acts at the center of pole.

Now, distance of center of pole from 'F₁' is given as:

d₂ = (L ÷ 2) - d₁

Now, as the pole is held horizontally straight, the moment about the point of application of force 'F₁' is zero for equilibrium of the pole.

So, Anticlockwise moment  = clockwise moment

Plug in the given values and solve for 'F₂'. This gives,

Therefore, the force exerted by the back arm on the pole is 58.8 N vertically down.

                                                                         

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.