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One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls sideways on the midpoint of the rope, displacing it a distance of 2 m (assume the car slides forward slightly in response to this, but then become stuck again). If he exerts a force of 87 N under these conditions, determine the force exerted on the car. Round your answer to one decimal place.

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lcmendozaf

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

[tex]L1^2=L2^2=L^2=2^2+(H/2)^2[/tex]  Replacing this value in the previous equation:

[tex]\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34[/tex]  Solving for H:

[tex]H=\sqrt{52}[/tex]

We can now, calculate the angle between L1 and the 2m segment:

[tex]\alpha = atan(\frac{H/2}{2})=60.98°[/tex]

If we make a sum of forces in the midpoint of the rope we get:

[tex]-2*T*cos(\alpha ) + F = 0[/tex]  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

[tex]T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N[/tex]

Ver imagen lcmendozaf
onyebuchinnaji

The force exerted on the car by the ropes is 372.8 N.

Angle formed when the rope is displaced

The angle formed when the rope is displaced is calculated as follows;

  • Let this angle = θ
  • The angle is between half of the rope and the car
  • Let the hypotenuse side = x

[tex](\frac{h}{2})^2 + 2^2 = x_1^2\\\\(\frac{h}{2})^2 + 2^2 = x_2^2\\\\\sqrt{(\frac{h}{2})^2 + 2^2} + \sqrt{(\frac{h}{2})^2 + 2^2} = 34\\\\2\sqrt{(\frac{h}{2})^2 + 2^2} = 34\\\\\sqrt{(\frac{h}{2})^2 + 2^2} = 17\\\\(\frac{h}{2})^2 + 2^2 = 289\\\\(\frac{h}{2})^2 = 285\\\\\frac{h}{2} = \sqrt{285} \\\\h = 2\sqrt{285} \\\\h = 33.76 \ m[/tex]

[tex]\theta = tan^{-1} (\frac{h/2}{d} )[/tex]

where;

  • L is length of the rope
  • d is the displacement

[tex]\theta = tan^{-1} (\frac{33.76}{4} )\\\\\theta = 83.2 ^o[/tex]

Tension on the ropes

The tension on each rope half of the rope is calculated as follows;

[tex]2Tcos\theta = F\\\\T = \frac{F}{2cos \theta} \\\\T = \frac{87}{2 \times cos(83.3)} \\\\T = 372.8 \ N[/tex]

Thus, the force exerted on the car by the ropes is 372.8 N.

Learn more about tension forces here: https://brainly.com/question/2008782

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