Exciting physics problem, anyone care for a challenge?

Question

An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope . The rope will break if the tension in it exceeds 2.45×104 N, and our hero's mass is 82.2 kg. The angle of the rope bent by the person in the middle to the horizontal holding it together forms a triangle.

If the angle between the rope and the horizontal is
theta = 11.9 degrees, find the tension in the rope.
Take the free fall acceleration to be g = 9.80 m/s^2.

What is the smallest value the angle (theta) can have if the rope is not to break?
Take the free fall acceleration to be g = 9.80 m/s^2.

Answer 1

Answerer #3 is close, but wrong.

When theta=11.9 degrees, the rope has not snapped and the man is at equilibrium (otherwise, the tension in the rope is 0). At equilibrium, all net forces are 0, by Newton's First Law. The man is assumed to be in the middle of the two cliffs, so the horizontal tension forces are symmetric, and hence cancel each other out. So, we only need to be concerned with vertical forces.

The downward force is only the force of gravity on the man, which is 82.2(9.8) N. The upward forces are the vertical components of the rope's tension on BOTH sides of the man: 2(Tsin(theta)).

So, 2Tsin(theta) - 82.2(9.8) = 0.

In the first question, to find tension, just plug for theta and solve for T.

In the second question, to find theta, just plug the maximum tension for T (=2.45x104N) and solve for theta.

Shame on answerers one and two for insulting the person who posted this question. This should be an open, supportive community. If you don't know, be quiet-- you just reveal your ignorance otherwise.

Answer 2

Since there are two ends of the rope, each supports half of the archaeologist's weight. This is the vertical component of the tensile force of the rope at one end. The horizontal component of the tensile force is determined by the angle of the rope with the horizontal, theta.
sin(theta) = (F/2)/T Eqn. (1)
Rearrange and solve for T, the tension in the rope
T = (F/2)/(sin(theta)).
Using F = 82.2 kg * 9.80 m/s^2 = 805.56 N and theta = 11.9 deg, the tension, T = 1953.3 N
For the smallest angle theta, we rearrange Eqn. 1 and solve for theta.
theta = asin ((F/2)/T)
where asin the arcsin or inverse sin function, sometimes written as sin^-1
Using the maximum tension given as 2.45*10^4N and F = 805.56N we find that 0.9086 degrees.

Answer 3

I think the tension in the rope is always

T = (82.2 * 9.8) / sin(theta)

I am encouraged to think my answer is correct, for my formula gives an increasing tension as theta gets smaller. The problem asks at what value of theta will the tension exceed a certain value. Make theta small enough and you can get the tension as big as you need.

Answer 4

You are the nerdiest person ever really you have nothing better to do than to challenge people in physics problems i think you should jus die for everyones sake

Answer 5

11.9

Answer 6

eat balls geek