Question
A 565 N mountain climber moves across a rope strung across a canyon. When she is more than halfway across, she notices the rope makes a 10 degree angle (from horizontal) behind her and a 25 degree angle (from horizontal) ahead of her. What are the tensions in the rope in front and behind her?
Answers
T1•sin10º+T2•sin25º=mg, ....(1)
T1•cos10º - T2•cos25º = 0.... (2)
From (2)
T1=T2•cos25º/ cos10º = 0.975•T2. ....(3)
Substitute (3) in (1)
0.174•0.975• T2+ 0.423•T2 = 565,
T2= 953.4 N
T1=929.5 N
T1•cos10º - T2•cos25º = 0.... (2)
From (2)
T1=T2•cos25º/ cos10º = 0.975•T2. ....(3)
Substitute (3) in (1)
0.174•0.975• T2+ 0.423•T2 = 565,
T2= 953.4 N
T1=929.5 N