A tightrope walker who weighs 640 N walks along a steel cable when he is halfway

Question

A tightrope walker who weighs 640 N walks along a steel cable when he is halfway across, the cable makes an angle of 0.040 rad below the horizontal.



U can just answer c)



a) What is the strain in the cable? Assume the cable is horizontal with a tension of 80 N before he steps onto it. Ignore the weight of the cable itself.

b) What is the tension in the cable when the tightrope walker is standing at the midpoint?

c) What is the cross-sectional area of the cable?

d) Has the cable been stretched beyond its elastic limit (2.5 x 108 Pa)?

Explanation / Answer

A) Solve the free-body diagram force balance to find the tension in the rope: Horizontal, not really gonna give more information. Vertical: W = 2*T*sin(theta) Solve for T: T = W/(2*sin(theta)) Find the total length of the cable when the tightrope walker is in the center: Trig relation: cos(theta) = L/(2*d) Solve for d: d = L/(2*cos(theta)) Strain in rope when loaded: epsilon = (2*d - L0)/L0 Strain in rope when pre-tensioned: epsilon0 = (L - L0)/L0 Linear elastic proportionality between loaded case and pre-tensioned case: T/T0 = epsilon/epsilon0 Summarize a system of equations to solve: T/T0 = epsilon/epsilon0 epsilon0 = (L - L0)/L0 epsilon = (2*d - L0)/L0 d = L/(2*cos(theta)) Solve for unknowns (epsilon, epsilon0, L0, and d): Assume L to be a known, even if it isn't. Result of solving (algebra not shown): L0 = L*( T*cos(theta) - T0)/(cos(theta)*(T - T0)) epsilon0 = T0*(1 - cos(theta))/(T*cos(theta) - T0) d = L/(2*cos(theta)) epsilon = T*(1 - cos(theta))/(T*cos(theta) - T0) Plug in expression for T, in to that for epsilon: epsilon = (W/(2*sin(theta)))*(1 - cos(theta)) / ((W/(2*sin(theta)))*cos(theta) - T0) Simplify: epsilon = W*(1 - cos(theta)) / (W*cos(theta) - 2*T0*sin(theta)) B) We already have an expression for tension: T = W/(2*sin(theta)) C) To find the cross sectional area, first find the stress expression: sigma = T/A Set up a Hooke's law equation: sigma = E*epsilon Substitute: T/A = E*epsilon Solve for A: A = T/(E*epsilon) Plug in expressions for T and epsilon: A = (W/(2*sin(theta))) / (E*(W*(1 - cos(theta))/(W*cos(theta) - 2*T0*sin(theta)))) Simplify: A = 1/2*(W*cos(theta) - 2*T0*sin(theta)) / (sin(theta)*E*(1 - cos(theta))) D: use our Hooke's law equation, updated with the previous result: sigma = 1/2*E*(W*cos(theta) - 2*T0*sin(theta)) / (sin(theta)*E*(1 - cos(theta))) Simplify: sigma = 1/2*(W*cos(theta) - 2*T0*sin(theta)) / sin(theta)/(1 - cos(theta)) --------------------------------- Summarize our resulting equations: A) epsilon = W*(1 - cos(theta)) / (W*cos(theta) - 2*T0*sin(theta)) B) T = W/(2*sin(theta)) C) A = 1/2*(W*cos(theta) - 2*T0*sin(theta)) / (sin(theta)*E*(1 - cos(theta))) D) sigma = 1/2*(W*cos(theta) - 2*T0*sin(theta)) / sin(theta)/(1 - cos(theta)) Data: E is the elastic modulus for steel. theta:=0.04 rad; W:=640 N; T0:=80 N; E:=200e9 Pa; Results: A) epsilon = 0.0008086 B) T = 8002 Newtons C) A = 0.00004948 m^2 or A = 49.48 mm^2 which corresponds to a diameter of D = 7.937 mm D) sigma = 9.898 MPa, which is much less than the elastic limit you've given

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