# Homework Help: Physics Help

1. Mar 9, 2005

2. Mar 9, 2005

### steve09090909

Michel (80kg) and Lauren (60kg) are spelunking in a cave. They are roped together with a 20 m long rope. Michel falls over a cliff.

A. If there is no friction, how long will it take lauren to also fall over the cliff.

B. She slid until 10m of the rope was over the edge of the pit, found her footing, and pulled Michel out. If she was able to get a coefficent of friction of 1.5 between her boots and the cave floor, what is the least amount of time that it would have taken her to pull him out.

3. Mar 10, 2005

### nd

Hi Steve,
A) Draw a diagram. There is only one force acting on Lauren. What is it? From this, what is her acceleration?
B) What is the maximum force Lauren can exert without slipping? This should correspond to maximum acceleration, and thus least time to pull Michel out.
Hope this helps.

4. Mar 10, 2005

### xanthym

Item A:
{Force From M's Weight} = FM = MM*g = (80 kg)*(9.81 m/sec^2) = (785 N)
The Force equation is {F = (mass)*(acceleration) = M*a}. The above computed Force "FM" is applied to the combination of M's + L's masses and they will accelerate together with constant acceleration "afall":
{Acceleration of M + L} = afall = FM/(MM + ML) =
= (785 N)/{(80 kg) + (60 kg)} = (5.61 m/sec^2)

The Distance "D" traveled under constant acceleration "afall" is given by:
D = D0 + v0*t + (1/2)*afall*t2
where D0 is the initial position (0 here), v0 the initial velocity (0 here), and "t" the elapsed time. The entire length of rope {D = 20 m} is traversed, so that using the value of acceleration "afall" computed above and solving for time "t":
(20 m) = (0) + (0) + (1/2)*(5.61 m/sec^2)*t2
::: ⇒ t2 = (7.13)
::: ⇒ t = (2.67 sec)

Item B:
{Coefficient of Friction} = K = (0.5)
{Distance L Slid before Gaining Footing} = s = (10 m)
{Distance from Cliff's Edge L Gained Footing} = X0 = (10 m)
{Maximum Pulling Force from L} = FL = K*{L's Weight} = K*{ML*g} =
= (1.5)*(60 kg)*(9.81 m/sec^2) = (883 N)
{Net Force} = F = FL - FM =
= (883 N) - (785 N) = (98 N)
The above Net Force "F" is applied to the combination of M's + L's masses, so the constant acceleration "apull" achieved is:
{Acceleration of M + L} = apull = F/(MM + ML) =
= (98 N)/{(80 kg) + (60 kg)} = (0.70 m/sec^2)

However, L slid distance s=(10 m) with initial constant acceleration "afall" during the time L had no footing, so L's velocity "v1" at the time L gained footing is determined from (where "v1" will be (-) since it's in the opposite direction that L is pulling):
(v1)2 - (v0)2 = 2*afall*s
::: ⇒ (v1)2 - (0) = 2*(5.61 m/sec^2)*(10 m)
::: ⇒ v1 = (-10.6 m/sec)

The Distance "X" from the cliff's edge is given by (where the edge is at X=0 and L is pulling in the +X direction):
X = X0 + v1*t + (1/2)*apull*t2
We again solve for "t" after placing appropriate values computed above into the equation. For this case, cliff's edge is at X=(0), L gained footing at X0=(10 m) and is pulling in the (+)X direction, and M will have been "pulled out" when L reaches X=(20 m):
(20 m) = (10 m) + (-10.6 m/sec)*t + (1/2)*(0.70 m/sec^2)*t2
::: ⇒ (0.35)*t2 - (10.6)*t - 10 = 0
::: ⇒ t = (31.2 sec)
Unfortunately, under the assumption that v1=(-10.6 m/sec) used above, L would be pulled over the cliff since L would reach X=(0) sooner than the above time and still have (-)velocity "v":
(0) = (10 m) + (-10.6 m/sec)*t + (1/2)*(0.70 m/sec^2)*t2
::: ⇒ (0.35)*t2 - (10.6)*t + 10 = 0
::: ⇒ t = (0.974 sec)
v = v1 + apull*t = (-10.6 m/sec) + (0.70 m/sec^2)*(0.974 sec) = (-9.92 m/sec)

We therefore assume the problem meant that L somehow achieved v1= (0 m/sec) at X0=(10 m), and re-solve for t:
(20 m) = (10 m) + (0)*t + (1/2)*(0.70 m/sec^2)*t2
::: ⇒ (0.35)*t2 - 10 = 0
::: ⇒ t = (5.35 sec)

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Last edited: Mar 10, 2005