Suppose an empty flatbed railroad car sits on a track as shown below, with a person pulling on a rope with a force of 310 Newtons in a direction of θ=25∘. Since the railroad car can't move in the direction that the person is pulling, not all of the force he is exerting works to move the car down the track. How much force is the person exerting toward the right (the force that would go towards trying to move the car)?  ____Newtons    The first person wasn't able to move the railroad car. Suppose a second rope is attached as shown below, and someone pulls on the rope in the direction β=18∘ south of east. If it takes 620 N to get the railroad car moving (even if only slightly), how much force must the second person exert in the given direction to get the railroad car to move? _____ Newtons​

Suppose an empty flatbed railroad car sits on a track as shown below with a person pulling on a rope with a force of 310 Newtons in a direction of θ25 Since the class=

Respuesta :

Force is a vector quantity, and a force can be resolved into its components to get the effect of the force.

The responses are;

  • a) The force exerted by the first person towards the right is approximately 280.96 Newtons.
  • b) To move the railroad car, the second person must exert a force of approximately 356.49 Newtons.

Reasons:

a) The force with which the person is pulling the railroad car, F = 310 N

The vector force on the rope [tex]\vec {v}_{rope}[/tex] = 310 × cos(25°)·i + 310 × sin(25°)·j

Which gives;

  • [tex]\vec {v}_{rope}[/tex] = 280.96·i + 131.01·j

The component of the vector acting along the railroad, towards the right is the i component

Therefore, the force the person exerts toward the right while trying to move the car is approximately

  • 280.96 Newtons

b) The direction of the rope of the second person, θ = 18°

The force required to get the railroad car moving = 620 N

The force the second person has to exert to the right is therefore;

F × cos(18°) = 620 - 310 × cos(25°)

Therefore;

[tex]\displaystyle F = \mathbf{\frac{620 - 310 \times cos(25^{\circ})}{ cos(18^{\circ})}} \approx 356.49[/tex]

The force the second person must exert in the given direction, is therefore;

  • F ≈ 356.49 Newtons

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