A 500-N tightrope walker stands at the center of the rope. If the rope can withstand a tension of 1400 N without breaking, what is the minimum angle the rope can make with the horizontal?

Answer 1

Hi Tomy!

The first step to almost every statics problem is to develop a free-body diagram. In this example, imagine that the tightrope walker can be analyzed as a point, and that, since he's walking on the center of the rope, the rope is deflecting symmetrically on both sides. Therefore, we have three forces acting on this point - the weight of the tightrope walker going directly down, and then the two tensions going up left and right at some angle with respect to the horizontal.

Our next step is to break down those tensions into component vectors. We can focus just on the Y-Directional forces here. The Y component of the tension force for each rope would be T*sin(theta), and it's in the positive direction. There are two of these forces directing upwards. The walker himself has a negative Y-Direction force of 500N. The tightrope walker must be in equilibrium - they're not falling through the rope, nor are they being accelerated upwards. Thus, the sum of forces is:

2*T*sin(theta) - 480 = 0 Newtons

The maximum tension in the rope is 1400N. If we substitute this in for T, we will get the critical angle at which the rope will break:

2*(1400)*sin(theta) - 480 = 0

Using algebra now, we can get an expression for theta:

theta = sin^-1(0.1714) = 9.87 degrees

Thus, the minimum angle the rope can make with the horizontal is 9.87 degrees. If you re-evaluate the equation in terms of theta, you can see what effect the angle has on the tension in the rope:

T = 240/sin(theta)

Note how shallower angles increase the tension, and as the angle approaches 0, the tension approaches infinity. You need a Y-Component of the tension to balance out the weight of the tightrope walker, and the more shallow the angle, the larger the tension must be to provide the proper component vector to balance the weight.

Hope this helps!