I cannot understand physics can someone please help me with this it something about finding force?

Question

A physicist finds that her car is stuck in the sand and cannot be driven out. Unable to push it out, she ties a strong rope from the front of the car to a large tree 30 m away and directly in front of the car. She then pushes in the middle of the rope with a force of 400 N in a direction perpendicular to the length of the rope. If the midpoint of the rope is displaced by 3.0 m, what is the force applied to the car?

Answer 1

Think of the rope pushed down like a triangle. Using an axiom of similar triangles we know that the ratio of the sides should equal the ratio of the forces. The physicist pushed it down 3.0m and the length of the rope from her to the car is 15m, which we'll make the hypotenuse. She pushes down at 400 N, so the hypotenuse is the answer you want.

F / 400 N = 15m / 3m

F / 400 N = 5

F = 2000 N

Bear in mind, that is the force applied along the rope. If you want the force being directed towards the tree, you will need to find the horizontal displacement and use that for the ratio. The horizontal displacement is easily found using Pythagorean's Theorem.

http://mathworld.wolfram.com/PythagoreanTheorem.html

Answer 2

Well, first of all you have to make a diagram of the situation. In this case it will be a triangle made by the car, the tree and the middle point of the rope that had a displacement of 3m.

You need to find the force the rope is applying to the car so you have to make some forces diagrams.

Now, you can have two situations: First, the rope is unelastic. That means that the rope won't have any defformation while the girl is pushing it. That means that the rope will allways measure 30m.

Second, the rope is elastic and it defforms when the girl is pushing it.

In the first case let's suppose the rope is not elastic and the car has started moving when the girl pushed it the 3 m. So you make your diagrams according to it.

Now you get a diagram where there is a vertical 400N force and two diagonal and equal forces in the opposite direction.

You just have to get one of the forces, let's call them Fa and Fb. Fa is the force applied to the car and Fb is the force applied to the tree.

Now all you have to do is calculate the angle between the car and the rope since for a rope, the force aplied goes right in the same direction the rope is going.

If the rope is not elastic, then you will have two triangles with a 15m hypotenuse.

in that case you get the angle with the sin function.

sin θ = 3/15

then θ=sin-1(3/15)

the angle will be:

θ = 11.537º

It will be the same angle in side A and in side B

now with this angle you can get the forces, remember that

ΣF=0 when the system is static, we are supposing it's static.

you can do it in many ways, by sepparating the forces into it's x and y componnets or by making a triangule with the forces vector, this is the method i'll use.

using sin law you get

400N/sin23.074º=Fa/sin 78.46º

solving for Fa you get:

Fa=400*(sin78.46º)/sin23.074º

so

Fa = 1000N and the same magnitude but different direction for Fb.

Now, let's suppose the rope is elastic, that would change the triangle because then you would get two rectangular triangles with one side measuring 3m and the other one measuring 15m, then you wouldn't know how much the hypotenuse measures.

so now the angle changes, in order to find the angle you use the tan function.

tan θ = 3/15

so

θ= tan-1(3/15)

θ = 11.31º

and you do the same to find the forces, but you use this angle now.

400N/sin(22.62º)=Fa/sin (78.69º)

solving for Fa you get

Fa = 400*sin(78.69º)/sin(22.62º)

Fa= 1019.8N

Hope this helps.

Answer 3

Distance between the car and tree is 30m. Depth of depression is 3m.

Weight in the middle of the rope is 400 N.

If T is the tension in each part of the rope, the vertical component is “T cos a”.

There are two “T cos a” that balance this 400 N.

2T cos a = 400 N

T cos a = 200 N

“T sin a” is the pulling force of the car.

The problem is to find the angle ‘a’.

The mid point of the rope is 15 m from the tree. The mid point is depressed to 3m vertically.

15 and 3 are the sides of a right angled triangle.

‘a’ is the angle between the side 3m and the hypotenuse of the triangle.

Tan a = 15/3 = 5;

The force that is pulling the car is T sin a

Let T sin a = X

T cos a = 200 N

Dividing the two,

tan a = 5 = X/ 200. Since tan a = 5 initially found.

X = 5 x 400 = 1000N.

Answer 4

Draw a free body diagram. Assuming the car is restricted to travel in a straight line towards the tree (the car would turn in real life), F_t_x applied to the car can be expressed as a function of the physicist's displacement. Geometrically, we note that she travels in a circular path around the tree, but she will be keeping her force vector constant.

Let theta be the angle between the line from the car to the tree and the rope.

tan(theta) = F_a_y / F_t_x
(since tension acts in the direction of the rope)
F_t_x = 400 / tan(theta)

we know sin(theta) = 3 / 15
theta = 11.537 deg

F_t_x = 400/ tan(11.537 deg)
F_t_x = 1.96 * 10^3 N --->Ans

Thus the full function is F_t_x = 400/ tan(asin(d/15))
where d is the y-displacement.
Note that theoretically, force on the car is unbounded at 0 deg.

bloomquist324, short answer, "no"...
slightly longer answer: Put down those equations before you hurt yourself; you sound like a kid who has stumbled across a set of scalpels...

pearlsawme, 15 is the hypoteneuse, not one of the legs, since the 30m is the rope length.

mensajeroscuro, why do you double the angle? My theta (11.527) is at the car....yes, we must assume inelastic rope.

Kookiemon's explaination is correct. I give the horz force (which is slightly less than the total), since the sprit of the question is to get the car out of the sand, and presumably, that's forward.

Answer 5

Well Michaelsgdec, thank you for having the correct answer. I really didn't think that insulting people who try to help was necessary, but since you did, let me just say Ouch! Drop the attitude. What possible purpose does it serve except your arrogance and ego? Obviously you are a brainiac because you are socially inept.