Problem 30 - College Physics

Question

In a pole-sliding game among friends, a \(90-\mathrm{kg}\) man makes a total vertical drop of \(7.0 \mathrm{~m}\) while gripping the pole which exerts and upward force (call it \(F_{\mathrm{p}}\) ) on him. Starting from rest and sliding with a constant acceleration, his slide takes 2.5 s. (a) Draw the man's free body diagram being sure to label all the forces. (b) What is the magnitude of the upward force exerted on the man by the pole? (c) A friend whose mass is only \(75 \mathrm{~kg}\), slides down the same distance, but the pole force is only \(80 \%\) of the force on his buddy. How long did the second person's slide take?

Step-by-Step Solution

Answer

(a) The forces are gravity downward and pole upward. (b) The upward force is 681.3 N. (c) The friend's slide takes 2.36 s.

Step 1: Draw the Free Body Diagram

To draw the free body diagram for the man sliding down the pole, identify all forces acting on him: the gravitational force \( F_g = m \cdot g \) acting downward, where \( m = 90 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \); and the upward force exerted by the pole \( F_p \). Thus, the diagram will show \( F_g \) pointing downward and \( F_p \) pointing upward.

Step 2: Calculate Gravitational Force

Calculate the gravitational force \( F_g \) acting on the man: \[ F_g = m \cdot g = 90 \cdot 9.81 = 882.9 \text{ N}. \]

Step 3: Determine Acceleration Using Kinematics

Use the kinematic equation to find the acceleration: \( s = ut + \frac{1}{2}at^2 \), where \( s = 7.0 \text{ m} \), initial velocity \( u = 0 \), and \( t = 2.5 \text{ s} \). Solving for \( a \), we get: \[ 7.0 = 0 + \frac{1}{2}a(2.5)^2 \implies a = \frac{7.0 \times 2}{2.5^2} = 2.24 \text{ m/s}^2. \]

Step 4: Apply Newton's Second Law

Apply Newton's second law: \( F_{net} = m \cdot a \), where \( F_{net} = F_g - F_p \). Using the man's mass and the acceleration: \( 90 \cdot 2.24 = 882.9 - F_p \).

Step 5: Solve for Upward Force \( F_p \)

Rearrange the formula to find \( F_p \): \[ F_p = 882.9 - (90 \times 2.24) = 882.9 - 201.6 = 681.3 \text{ N}. \]

Step 6: Determine Force on Friend

The force on the friend's slide is \( 80\% \) of the force on the man: \( 0.8 \times 681.3 \text{ N} = 545.04 \text{ N}. \)

Step 7: Calculate Friend’s Acceleration

Using the net force equation for the friend: \( F_g = 75 \cdot 9.81 = 735.75 \text{ N} \), so \( F_{net} = 735.75 - 545.04 \), acceleration \( a = \frac{F_{net}}{m} = \frac{735.75 - 545.04}{75} = 2.546 \text{ m/s}^2 \).

Step 8: Compute Time for Friend’s Slide

Rearrange the kinematics equation for time \( t \): \( 7.0 = 0 + \frac{1}{2} \times 2.546 \times t^2 \). Solving for \( t \) yields: \[ t = \sqrt{\frac{7.0 \times 2}{2.546}} = 2.36 \text{ s}. \]

Key Concepts

Free Body Diagram

A free body diagram is an essential tool in physics for visualizing the forces acting on an object. In the case of the pole-sliding man, we have two primary forces to consider:

• The gravitational force (weight), which acts downward, pulling the man towards the Earth.
• The upward force exerted by the pole as he slides, which opposes the gravitational force.

In drawing this diagram, you start by sketching the man as a point or a small box. Use arrows to represent the direction of each force. The length of the arrow indicates the magnitude of the force:

• Label one arrow as the gravitational force, denoted as \( F_g \), calculated by the product of mass and gravitational acceleration \( m \cdot g \).
• Label the opposite arrow as the pole's force \( F_p \), which you will calculate later.

Ensure the direction and label of each force are clear, forming a foundation for further analysis.

Kinematics

Kinematics involves the study of motion without considering the forces causing it. In our exercise, we delve into this concept by analyzing the motion of the man sliding down the pole.
He starts from rest, which implies his initial velocity (\( u \)) is zero. We employ one of the core kinematic equations to find the acceleration (\( a \)) of his motion:

• \[ s = ut + \frac{1}{2}at^2 \]
• Where \( s \) is the displacement (7.0 m), \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time (2.5 s).
• Rearranging for \( a \), the formula becomes \[ a = \frac{2s}{t^2} \].

This equation helps us determine how quickly the man speeds up (or down the pole, in his specific scenario). By determining \( a \), we can better understand other dynamics at play.

Gravitational Force

Gravitational force is a fundamental force that attracts two bodies towards each other. The Earth's gravitational pull is what pulls objects towards its center. In the context of the exercise:

• The gravitational force exerted on the man is calculated using \( F_g = m \cdot g \).
• Where \( m \) is the mass (90 kg) and \( g \) is the gravitational acceleration (approximately \( 9.81 \, \text{m/s}^2 \) on Earth).
• Thus, \( F_g = 90 \times 9.81 = 882.9 \, \text{N} \).

This force acts downward and is a constant presence, serving as a crucial factor in the motion and forces of the problem. Understanding gravitational force allows us to predict how much force must be applied to counteract or supplement its effects in various scenarios.

Acceleration

Acceleration is the rate of change of velocity of an object. When tackling this problem, finding the acceleration is imperative to understand how the man's speed changes as he descends the pole.

• Initially, the man starts from rest, meaning his initial velocity is zero.
• Using the kinematic equation, we've calculated the acceleration to be approximately \( 2.24 \, \text{m/s}^2 \).
• This indicates how quickly the man reaches his final speed during the 2.5-second descent.

Acceleration is influenced by the net force acting on an object. Understanding this relationship allows us to solve for unknown forces when examining free-fall motion or similar dynamics.

Force Calculation

Calculating force is crucial to understand interactions in physics. Newton's Second Law of Motion, \( F_{net} = m \cdot a \), serves as the basis for calculating net force and individual forces acting on an object.

• For the man sliding down the pole, the net force is the difference between the gravitational force and the upward force from the pole.
• Using the man's mass and calculated acceleration: \( F_{net} = 90 \times 2.24 = 201.6 \, \text{N} \).
• The upward force, \( F_p \), can be calculated by rearranging the formula to \( F_p = F_g - F_{net} \).

By solving for each force, we get not only the magnitude of each but a clearer picture of the interaction between the forces involved, particularly how the pole's supportive force helps oppose the man's gravitational pull.