Question

You (85 kg) are standing on the horizontal surface at the top of a cliff. The coefficient of static friction between your feet and the ground is .5 and the kinetic coefficient is .4. You have a friend (100kg) that wants to accelerate down the side of the cliff. They set up massless and frictionless pulley and secure themselves with a massless string (physics students can do these sorts of things) that is then tied around your waist. The coefficient of static friction between your feet and the ground is .5 and the kinetic coefficient of friction is .4.

Answer 1

a = 7.6 m / s²

Step-by-step explanation:

In this exercise we must use Newton's second law . In general, it is asked to know if the system is in equilibrium or accelerated, suppose that the system is in equilibrium

Person on top of the cliff m = 85 kg

Y axis

N - W = 0

N = w₁

N = m g

X axis

T’ -fr = 0

T’ = fr

the friction force has the formula

fr = μ N

we subtitle

T’ = μ m g

we calculate

T ’= 0.5 85 9.8

T’ = 416.5 N

This is the maximum value that the tension of the rope can have by remaining in static equilibrium.

Person on the cliff M = 100 kg

T -W = 0

T = W = M g

T = 100 9.8

T = 980 N

we see that

T> T ’

as the tension exceeds the maximum static friction force, in the system it is accelerated, so the friction coefficient decreases to μ= 0.4

We medo the problem, but with acceleration

Person on the cliff

T ’-fr = m a

a = (T '- μ mg) / m

in this case, how the rope should maintain the tension of the diagram of the person hanging

T ’= T = Mg

we substitute

a = Mg / m - very g

a = g (M / m - my)

let's calculate

a = 9.8 (100/85 - 0.4)

a = 7.6 m / s²