https://session.masteringphysics.com/myct/itemview7offset-nex UPEI Phys 111E- 20

Question

https://session.masteringphysics.com/myct/itemview7offset-nex UPEI Phys 111E- 2015 columbla Unv Exercise 5.21 Columbia Univ Part A Inna K The climbers, roped together, are sliding freely down an icy mountainside. The upper climber (mass 80 kg) is on a slope at 12 to the Express your answer using two significant figures See more thorizontal, but the lower climber (mass 86 kg ) has gone over the edge to a steeper slope at 38 l a= 3.8 m/s2 Answer Requested VE YOURS GIFT OF MO I your old books safely and eas complete stop, what force must the ax exert against the ioe? F 304 Give Up Incorrect; Try Again;5 attempts remaining Continue

Explanation / Answer

Find the acceleration of the system, then use F = ma


................W = 80kg(9.8)
.................
.......................................
................T(sin12°).........F = unbalanced force
..........T
......
.._12° ____________


F = W + T(sin12°)
F = ma
W + T(sin12°) = ma
T(sin12°) = ma - W
T(sin12°) = (80kg)a - (80kg)(9.8)..........<==eq1

....T(sin32°)
.'...
.....|
.....|....T
.....|../....
.....|./..32°.
....--------.
.....W = (65kg)(9.8)
..

Unbalanced force F = W - T (sin38°)
F = ma
W - T (sin38°) = ma
T (sin38°) = W - ma
T (sin38°) = (65kg)9.8 - (65kg)a<===eq2

divide equation 1 by equation 2

T(sin12°) ......... (80kg)a - (80kg)(9.8)..........<==eq1
=
T (sin36°) .........(65kg)9.8 - (65kg)a............<===eq2

.................. (80kg)a - (80kg)(9.8)....
0.352 =
........... ......(65kg)9.8 - (65kg)a......

0.352(637 - 65a ) = 80a - 784
224.224 - 22.88a = 80a - 784
1008.224 = 102.88a

a = 9.8m/sec^2 downwards

(they can't have an acceleration greater than g, free fall)

The upper climber made the axe exert a force equal to

F = ma
...= 80kg(9.85)N
F = 788N

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