Two ice climbers, Nana and Popo, are walking up a steep ice slope with an incline angle of 40 degrees. They are
simply walking up it with ice spikes. The ice spikes give each of them a coefficient of kinetic friction of μk=0.7 as long as they are standing. The climbers are attached by a long length of rope. Popo trips, falls, and begins sliding down the slope. Unfortunately, there is a shear drop off not far behind them. Popo goes over the edge.
A) When the rope goes taut, if both climbers have masses of 60 kg for Nana and 80 kg for Popo, what is Nana’s acceleration as she is pulled towards the cliff (she keeps her ice spikes dug into the ground)?
B) While sliding, Nana builds up a speed of 5 m/s before pulling out her climbing axe and burying it into the hillside! This increases her total coefficient of kinetic friction of μk=2.6. What is her acceleration now?
C) How far down the slope does she slide after using the axe?
D) Nana manages to stop their decent. While stopped, with the axe in the ice, her coefficient of static friction is μs=2.8. What is the maximum acceleration ice Popo can climb up the rope, without causing them to slide again?
Denote Nana's mass as `m_1` and Popo's mass as `m_2.` Denote the incline angle as `alpha.`
Draw a force diagram for Nana. There are 4 forces: its own weight `m_1*g,` normal force `N` , friction force `F_f` and traction force from Popo `F_T.` By Newton's Second Law the sum of these forces is equal to `m_1*a,` where `a` is her acceleration.
Consider axes x and y as shown on the picture. No doubt that acceleration `a` is parallel to the x-axis. Project forces on x and y.
So we obtain
[Compute `m_2/m_1+sin(alpha) approx 1.976` , `cos(alpha) approx 0.766` ]
A) `mu_1=0.7,` `a_1=g*(0.7*0.766-1.976)=-g*1.440=-14.110 (m/s^2).`
(minus means to the left, i.e. downward)
B) `mu_2=2.6,` `a_1=g*(2.6*0.766-1.976)=g*0.0157=0.154(m/s^2).`
Now acceleration is upward, i.e. Nana brakes (slows) but very slowly.
C) The speed of Nana is `V_1-t*a_2,` it becomes zero at `t_1=V_1/a_2.`
The distance is `V_1*t_1-(a_2*t_1^2)/2=(V_1^2)/(2a_2)=25/0.308 approx 81(m).`
Hope there is enough space.
D) `mu_3=2.8` and Popo's acceleration is `a_3,` so `F_T=m_2*(g+a_3)`
(or `m_2` transforms to `m_2*(1+a_3/g)` ).
The maximum `a_3` is found from
`a_3=g*[(m_1/m_2)*(mu_3*cos(alpha)-sin(alpha)) - 1] = `
`=g*[3/4*(2.8*0.766-0.643) - 1]=g*0.126=1.24 (m/s^2).`
I made a mistake when supposed that `F_(T)=m_2g.` This is incorrect.
The correct way is to consider Popo and forces acting on him.
One of them is `m_2g` downwards and another is `F_T` upwards (the same magnitude that pulls Nana downhill). And the net force is `m_2a` (Nana and Popo have the same magnitude of the acceleration). Popo's acceleration `a` is directed downwards, so to preserve the sign of a, we have to use vertical y-axis for Popo:
Now consider again
and substitute `F_(T)=m_2(g+a)` into it:
From this equation
Now for the answers.
A) `mu_1=0.7,` `a_1=9.8*(0.7*60*0.766-60*0.643-80)/140 approx -6.05 (m/s^2)`
(minus means to the left, i.e. downward).
B) `mu_2=2.6, a_2=9.8*(2.6*60*0.766-60*0.643-80)/140 approx 0.064 (m/s^2).`
Even slower than I computed first...
C) the distance is `25/(2a_2)=25/0.128=195.3 (m).`
D) Was correct above.