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You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35cm from its tip.How much force must you exert to keep the pole motionless in a horizontal position?NExpress your answer using two significant figures.

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cjmejiab

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as [tex]F_2[/tex] and the Tension in fence post as [tex]F_1[/tex], then

[tex]\sum F = 24(9.8)[/tex]

[tex]F_1 + F_2= 24(9.8)[/tex]

We apply summatory of moments then

[tex]F_2*1.25 = F_1*1.6[/tex]

Where the Force 2 is 1.25m from the center of summatory,

We can note that,

[tex]1.6 m - 0.35m=1.25m[/tex]

We have two equation and two incognites, then replacing (1) in (2)

[tex]1.6(235.2 -F_2) = 1.25F_2[/tex]

[tex]376.32 = F2(1.6+1.25)[/tex]

[tex]F_2= \frac{376.32}{2.85}[/tex]

[tex]F_2 =132 N[/tex]

onyebuchinnaji

The force you must exert to keep the pole is horizontal position is 132 N.

The given parameters;

  • length of pole, l = 3.2 m
  • mass of the pole, m = 24 kg

The weight of the pole is calculated as;

W = 24 x 9.8 = 235.2 N

The sum of the forces exerted to keep the pole horizontal;

[tex]\Sigma F= F_1 + F_2 = 235.2 \\\\F_1 = 235.2 - F_2[/tex]

Take moment about the center of the pole;

[tex]F_1 \times 1.6 = F_2 (1.6 - 0.35)\\\\1.6(235.2 - F_2) = 1.25F_2\\\\376.32 - 1.6F_2 = 1.25 F_2\\\\376.32 = 2.85F_2\\\\F_2 = \frac{376.32}{2.85} \\\\F_2= 132 \ N[/tex]

Thus, the force you must exert to keep the pole is horizontal position is 132 N.

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