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Consider the 52.0 kg mountain climber in the figure. 15° (a) Find the tension in the rope in N) and the force that the mountain climber must exert with her feet in N) on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. tension in rope force on feet ON (b) What is the minimum coefficient of friction between her shoes and the cliff?

Respuesta :

a. the force that the mountain climber must exert with her feet in N) on the vertical rock face to remain stationary is 273 N.

b. the minimum coefficient of friction between her shoes and the cliff is 0.268.

What is the term of friction?

The force that prohibits one solid object from traveling across another is known as friction. Static friction, sliding friction, rolling friction, and fluid friction are the four main categories of friction.

a).

Balancing the forces in x-direction,

F cos(15) = T sin(31)

F = T sin(31)/cos(15)     ...... (1)

balancing the forces in y-direction,

T cos(31) + F sin(15) = 52 x 9.8 = 509.6

T cos(31) + (T sin(31)/cos(15)) sin(15) = 509.6

T = (509.6 cos(15))/(cos(31) cos(15) + sin(31) sin(15))

T = 512 N

Force, F = 512 sin(31)/cos(15)

= 273 N

b)

Coefficient of friction,

u = sin(15)/cos(15)

= 0.268

To learn more about friction from the given link.

https://brainly.com/question/24338873

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