Determine the magnitudes of the forces she must exert with each hand. Determine

Question

Determine the magnitudes of the forces she must exert with each hand.

Determine the directiones of the forces she must exert with each hand.
a) both and are directed downward
b)is directed downward, is directed upward
c) is directed upward, is directed downward
d) both and are directed upward

To what position should she move her left hand so that neither hand has to exert a force greater than 190 ?


To what position should she move her left hand so that neither hand has to exert a force greater than 85 ?

Explanation / Answer

We need some units and more information or a clear diagram to fully solve this.
She could just hold the pole with one hand and not exert any force on it with the other hand, but let's assume that she is only applying vertical forces and not any torques. In that case, she would push down with her hand that is on the end of the pole and push up with the hand that is closer to the middle. So, either choice b or choice c depending on how the hands are labeled in the problem.

starting at her right hand looking at torque her left hand must cancel out the torque from the pole. So force times distance to her left hand must cancel pole weight times location of the center of mass of the pole from her right hand
Frh (0.0m) + Flh (0.3m) + W (1.5m/2) = 0 (right hand contributes nothing since distance is zero)

Flh = 9.5 (9.8) (.75) / 0.3 = 233N or better, 230 N

and now the up and down forces

Frh + W + Flh = 0

Frh = - Flh - W = -233N +93.1N = -139.65 N (down) or better, 140 N down

Middle part

If we move the left hand, the force exerted by the right hand will also decrease, so lets set

Flh = 190 and see where location is

dlh = W(1.5m/2)/Flh = 9.5 (9.8) (.75) / 190 = .3675m or .37m away from right hand (which is still at end of pole)

check this .37 (190) = 9.5 (9.8) (.75) ? yes

check right hand force 190 - 93.1 < 190 yes force is less than required

Last part

Let's move the left hand and afterwards see if right hand is below the required amount.

Flh = 85

dlh = 9.5 (9.8) .75 / 85 = 0.82 m

what does this give for the right hand?

Frh = + 85 - 93.1 N = - 8.1 N in direction of Flh So now that the left hand is past the mid point, both hands are pushing UP on the pole

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