Question

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs W = mg = 520 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the harness are not the same.

(A.) Draw the three forces acting on the harness. Set up a coordinate system and break forces into their vertical and horizontal components.

(B.) Write Newton's Second Law in component form for the vertical and horizontal forces on the climber. What are the accelerations, a_x and a_y?

(C.) What are the magnitudes of the tensions in the two ropes, T₁and T₂? [A useful Trig identity: sin(?+?) = sin ? cos ? + cos ? sin ?]

Answer 1

Weight of the person, W 520 N The diagram showing the force acting on the hamess is given below, TL TR 65.0 go.o° 80.0° B) x- component of force, E=T2 sin 80°-Tsin 65。 max = T2 sin 800-Tsin 650

a, =-(T, sin 80°-T sin 65° ) (0.9857,-0.9063 Z) 53.06 0.0 1867-0.017T (1) y- component of force, TF,, T, cos 800 + T cos 650-W may=T2 cos 80°+Teos 65°-520 ay (T, cos 800 + T1 cos 650-520) _(0.1736T, +0.4226T,-520) 53.06 3.27 x10-3T2 +7.96 x10-3-9.8 (2) C)

Consider the equations, T, sin 80-Tn ..3) 7, cos 80% T1 cos 65°-W-0 7, cos80%Teos 650-r (4) T^ sin 80°-T, sin 65° I^ sin 65° sin 80 -0.927, (5) Substituting (3) in (2), 520 0.92T cos 80° +cos 650 520 0.167 +0.42267 520 = 0.5826T, Hence tension

520 1 . . (1582廓 1892.5ー Substituting, 7, 092×T, = 0.92 x 892.5 -1821-12 N Substituting the results in (1) and (2), a,-0.01867-0.0 17T 0.0186×821.12-0.0 17 × 892.5 0.1m/s a, 3.27x10-3T, +7.96x109 T-9.8 -3.27x821.12x103 +7.96x892.5 x10-3-9.8 -0.01m/s