Problem 72 - College Physics

Question

A person has a choice while trying to push a crate across a horizontal pad of concrete: push it at a downward angle of \(30^{\circ},\) or pull it at an upward angle of \(30^{\circ} .\) (a) Which choice is most likely to require less force on the part of the person: (1) pushing at a downward angle; (2) pulling at the same angle, but upward; or (3) pushing or pulling shouldn't matter? (b) If the crate has a mass of \(50.0 \mathrm{~kg}\) and the coefficient of kinetic friction between it and the concrete is \(0.750,\) calculate the required force to move it across the concrete at a steady speed for both situations.

Step-by-Step Solution

Answer

Pulling upward requires less force because it reduces the normal force and thus the friction.

Step 1: Analyze the Force Components

For both pushing and pulling, the force can be split into horizontal (parallel) and vertical (perpendicular) components. When pushing downward, the vertical component increases the normal force, whereas when pulling upward, it decreases the normal force. The normal force affects the kinetic friction.

Step 2: Calculate Normal Force for Pushing

When pushing at a downward angle, the normal force (N) becomes \(N = mg + F \sin(\theta)\), where \(m = 50.0\) kg is the mass of the crate, \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity, \(F\) is the applied force, and \(\theta = 30^{\circ}\).

Step 3: Calculate Normal Force for Pulling

When pulling at an upward angle, the normal force becomes \(N = mg - F \sin(\theta)\), reducing the effective normal force acting on the crate.

Step 4: Set Up Force Equations for Horizontal Movement

For steady speed, the horizontal component of the force (\(F \cos(\theta)\)) equals the frictional force \(f_k = \mu N \), where \(\mu = 0.750\) is the coefficient of kinetic friction. This gives the equation \(F \cos(\theta) = \mu N\) for both situations but with different expressions for \(N\).

Step 5: Solve Equations for Pushing

Substitute \(N = mg + F \sin(\theta)\) into \(F \cos(\theta) = \mu N\): \(F \cos(30^{\circ}) = 0.750(mg + F \sin(30^{\circ}))\). Solve for \(F\).

Step 6: Solve Equations for Pulling

Substitute \(N = mg - F \sin(\theta)\) into \(F \cos(\theta) = \mu N\): \(F \cos(30^{\circ}) = 0.750(mg - F \sin(30^{\circ}))\). Solve for \(F\).

Step 7: Comparison of Forces

Compare the magnitude of the forces calculated for pushing and pulling to determine which requires less effort.

Key Concepts

Pushing vs Pulling

In physics, understanding the difference between pushing and pulling an object at an angle is essential for determining the effort required to move that object. When you push a crate downward at an angle, you're not only moving it forward but also pressing it more into the ground, increasing the normal force. On the other hand, when you pull the crate upward, part of your force lifts the crate, decreasing the normal force.

This change in the normal force is crucial because it directly affects the frictional force acting against the motion of the crate. Since the frictional force is proportional to the normal force, pulling upward often requires less force than pushing downward at the same angle. This is because the normal force, and hence the friction, is reduced when pulling compared to pushing.

• Pushing increases normal force and friction.
• Pulling decreases normal force and friction.
• The difference affects the needed applied force to move the object.

Normal Force

Normal force, often denoted by \(N\), is a crucial factor in the dynamics of objects on a surface. It's the force exerted by a surface to support the weight of an object resting on it, and it acts perpendicular to the surface. The normal force adjusts according to the components of external forces applied to the object.

When an external force is applied at an angle, it can alter the magnitude of the normal force. For instance, when pushing down on an object at an angle, the vertical component of the force increases the normal force, while pulling the object up decreases the normal force.

• Normal force acts perpendicular to the contact surface.
• Pushing down increases, while pulling up decreases the normal force.
• Normal force affects the frictional force directly.

Kinetic Friction

Kinetic friction is the resistance that occurs when two objects slide against each other. It plays a significant role in how much force is needed to maintain motion at a constant velocity. The kinetic frictional force \(f_k\) is calculated using the formula \(f_k = \mu N\), where \(\mu\) is the coefficient of kinetic friction and \(N\) is the normal force.

In our scenario of moving a crate across concrete, the kinetic friction depends heavily on the normal force, which varies with pushing or pulling. More specifically:

• Higher normal force results in higher kinetic friction.
• Pushing a crate increases kinetic friction due to added normal force.
• Pulling reduces kinetic friction as the normal force decreases.

Knowing the kinetic friction is crucial for calculating the precise amount of applied force required to keep the object moving steadily.

Force Components

Whenever a force is applied at an angle, it can be split into more manageable components: horizontal and vertical. Understanding these components is key to solving force and motion problems.

The horizontal component of the force is responsible for moving the object along the surface. In both pushing and pulling, this horizontal component can be expressed as \(F \cos(\theta)\), where \(\theta\) is the angle of the applied force. This component must overcome the friction to maintain constant speed.

• Horizontal component: \(F \cos(\theta)\) - moves the object along the surface.
• Vertical component: \(F \sin(\theta)\) - affects the normal force.
• Solving force problems relies on correctly separating and using these components.

Comprehending these force vectors allows students to analyze and predict the physical behavior of objects in motion under various external forces.