Physics Word Problem?

Question

2 teams of childeren playing tug of war are well matched, and they pull on the rope with exactly the same force. They have placed a 65N weight in the center of the rope. The teams first pull with 750N of force each, then in a heroic effort to straighten the rope, pull with 1,500N of force.

What is the angle of the rope makes with the horizontal with

A) a force of 750N on each end of the rope?
B) a force of 1500N on each end.

I am having a hard time understanding this one...so if you could provide details, that would help a lot. One thing is that the children are applying force in opposite directions...so the total for does not = double one side, right? Or because the force is placed on the 65N weight----do you add the force from each end. Anyway details would help. thanks

Answer 1

You can also use trigonometry
- the weight of 65N provides the vertical component
- the pull of the rope (750/1500) provides the hypotenuese

Since Hypotenuese = 750N and Opposite = 65N
If the angle is Q

Then sin Q = Opposite/Hypotenuse

You can then solve for Q by calculating Opposite/Hypotenuse and then inversing sin it.

I don't think the force will be doubled as the forces act in opposing directions, hence the angle on both sides will be the same...i think...

Answer 2

Here's one way to look at it:

Team "A" (let's say they're on the left side) is pulling leftward on the rope with a force of 750 N.

A-------O-------B

That means the rope is pulling rightward on Team "A" with 750N (Newton's 3rd Law)

That means the tension in the rope is 750N (the tension always equals the amount of force that the rope is exerting).

That means the rope (the left-hand rope) is pulling leftward with 750N on the object "O".

By similar reasoning, the right-hand rope is pulling rightward with 750N on the object.

Implicit in the problem (though not explicitly stated) is that the weight of the object causes the system to sag in the middle. So the ropes are not really horizontal--they form a wide "V" shape with the object at the bottom of the V. The ropes are pointing somewhat diagonally.

Now to figure out the angles of the ropes, look at the forces acting on the center object. There are 3 of them:

* Force of gravity (65N straight down);
* Tension from the left rope (750N upward and to the left);
* Tension from the right rope (750N upward and to the right).

Now, since the object is not accelerating, those 3 force vectors must balance out. Decompose the two tension vectors into their horizontal and vertical components (hint: this will involve the unknown angle "a" that the ropes make with the horizontal). Then write equations that make all the horizontal forces cancel out and all the vertical forces cancel out. That should be enough to let you solve for angle "a".

For the 2nd part (tension = 1500N), it's solved exactly the same way, but you just change the the value of the tension.

Answer 3

The rope tension, T, is the hypotenuse; the weight, W, is the side opposite. So the sine of the angle is W/T. Therefore, the angle is the arcsin of W/T.

Answer 4

You need to draw a scale diagram, with vectors representing the forces.