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Mountain climbers with masses \(m\) and \(M\) are roped together while crossing a horizontal glacier when a vertical crevasse opens up under the climber with mass \(M\). The climber with mass \(m\) drops down on the snow and tries to stop by digging into the snow with the pick of an ice ax. Alas, this story does not have a happy ending, because this doesn't provide enough friction to stop. Both \(m\) and \(M\) continue accelerating, with \(M\) dropping down into the crevasse and \(m\) being dragged across the snow, slowed only by the kinetic friction with coefficient \(\mu_{k}\) acting between the ax and the snow. There is no significant friction between the rope and the lip of the crevasse. (a) Find the acceleration \(a\).(answer check available at lightandmatter.com) (b) Check the units of your result. (c) Check the dependence of your equation on the variables. That means that for each variable, you should determine what its effect on \(a\) should be physically, and then what your answer from part a says its effect would be mathematically.

Step by step solution

01

Set up the forces acting on both climbers

Consider the forces acting on the climber of mass \( m \) and the climber of mass \( M \). The force of gravity on the climber of mass \( M \) is \( Mg \), where \( g \) is the acceleration due to gravity. The tension in the rope is \( T \), and the force of kinetic friction acting on the lighter climber is \( f_{k} = \mu_{k}mg \).

02

Apply Newton's second law to the heavier climber

For the climber with mass \( M \):\[ Mg - T = Ma \]where \( a \) is the acceleration of the system. Rearranging gives:\[ T = M(g - a) \]

03

Apply Newton's second law to the lighter climber

For the climber with mass \( m \):\[ T - f_{k} = ma \]Substitute \( f_{k} = \mu_{k}mg \) and \( T = M(g - a) \) from the previous step:\[ M(g - a) - \mu_{k}mg = ma \]

04

Solve for the acceleration \( a \)

Rearrange the equation from Step 3:\[ M(g - a) - \mu_{k}mg = ma \]Simplifying, we get:\[ Mg - Ma - \mu_{k}mg = ma \]\[ (M + m)a = Mg - \mu_{k}mg \]Thus, solving for \( a \):\[ a = \frac{M - \mu_{k}m}{M+m} g \]

05

Check the units of \( a \)

The units of acceleration are meters per second squared \( \text{m/s}^2 \). Both the numerator \( (M-\mu_{k}m)g \) and the denominator \((M+m)\) have units that cancel appropriately, leaving \( g \), which is \( \text{m/s}^2 \). Thus, \( a \) has the correct units.

06

Analyze the dependence on variables

Physically, if \( M \) increases, the acceleration \( a \) increases since more mass accentuates the gravitational effect on \( M \). Mathematically, as \( M \) increases, so does \( M - \mu_{k}m \) in the numerator, increasing \( a \). If \( \mu_{k} \) increases, \( a \) decreases since friction counteracts motion, matching the mathematical behavior since \( \mu_{k}m \) increases the negative term in the numerator. Increasing \( m \) would increase the denominator more than the negative term in the numerator, thus reducing \( a \), which is consistent with the physical understanding.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Friction

Friction is the force that opposes the relative motion between two surfaces in contact. It's a vital force that we encounter daily, from walking to writing. In our exercise scenario, friction acts as a resisting force that opposes the motion of the climber with mass \( m \) who is being dragged across the snow by the climber with mass \( M \). This type of friction is kinetic friction, which applies when two objects are moving relative to each other.

Friction depends on two main factors:

• The types of materials in contact: Different interactions between surfaces determine how much they "stick" to each other.
• The normal force: This is the force perpendicular to the surfaces in contact, often associated with the object's weight when it's on a horizontal surface.

The kinetic friction force can be calculated using the formula \( f_{k} = \mu_{k} N \), where \( \mu_{k} \) is the kinetic friction coefficient and \( N \) is the normal force.

In our exercise, the kinetic friction force is calculated as \( f_{k} = \mu_{k}mg \), with \( mg \) being the normal force provided by the weight of the climber of mass \( m \). Friction ultimately counteracts the motion caused by the gravitational pull on the heavier climber, but in this case, it is not sufficient to prevent the lighter climber from being pulled along.

Acceleration

Acceleration is defined as the rate of change of velocity with time. It is a vector quantity, which means it has both magnitude and direction. In the context of Newton's second law, acceleration is directly proportional to the net force acting on an object and inversely proportional to its mass: \( a = \frac{F_{ ext{net}}}{m} \).

In this problem, acceleration \( a \) affects both climbers as they move due to the combined forces acting on each. The force of gravity pulls the heavier climber downward with a force \( Mg \), while both the tension in the rope and kinetic friction force oppose the motion. The effective net force is distributed over the total mass of the system \( (M+m) \), resulting in an overall acceleration given as\[a = \frac{M - \mu_{k}m}{M+m} g\]

This equation illustrates how different factors contribute to acceleration:

• Higher mass \( M \) of the dropped climber increases gravitational influence, thus increasing \( a \).
• Higher friction coefficient \( \mu_{k} \) serves to slow down, reducing \( a \).
• The mass of the other climber \( m \) affects the system in a complex interplay with friction.

Understanding acceleration in this context helps us see how various forces resultantly influence the motion of objects in different scenarios.

Kinetic Friction Coefficient

The kinetic friction coefficient \( \mu_{k} \) is a dimensionless constant that represents the ratio of the force of kinetic friction between two surfaces to the normal force pressing them together. Essentially, it tells us how "sticky" two surfaces are while in motion relative to each other.

This coefficient depends on the materials and the condition of the surfaces in contact. For instance, ice and steel in contact would likely have a lower \( \mu_{k} \) than rubber on concrete.

• A low \( \mu_{k} \) means the surfaces slide easily over one another, meaning less force is resisted by friction.
• A high \( \mu_{k} \) indicates greater resistance, leading to more force required to keep sliding.

In the given scenario, \( \mu_{k} \) represents the frictional resistance between the climber's ice axe and the snow. The kinetic friction coefficient here directly impacts the deceleration of the climber with mass \( m \).

Mathematically, we see that as \( \mu_{k} \) increases, the negative term \( \mu_{k}mg \) in the force balance equation increases, effectively reducing the acceleration \( a \) as calculated.