Problem 78 - College Physics

Question

An object (mass \(3.0 \mathrm{~kg}\) ) slides upward on a vertical wall at constant velocity when a force \(F\) of \(60 \mathrm{~N}\) acts on it at an angle of \(60^{\circ}\) to the horizontal. (a) Draw the freebody diagram of the object. (b) Using Newton's laws find the normal force on the object. (c) Determine the force of kinetic friction on the object.

Step-by-Step Solution

Answer

Normal force: 30 N, Kinetic friction: 22.56 N.

Step 1: Understand the given information

We need to analyze a scenario where a box of mass 3.0 kg is sliding up a vertical wall with constant velocity. A force \( F = 60 \text{ N} \) is applied at an angle of \( 60^{\circ} \) to the horizontal.

Step 2: Free Body Diagram

In the free body diagram, we identify the forces acting on the object:1. The gravitational force \( \vec{F}_g \) acting downward, \( F_g = mg = 3.0 \times 9.8 = 29.4 \text{ N} \).2. The applied force \( \vec{F} \) with components \( F \cos 60^{\circ} \) horizontally and \( F \sin 60^{\circ} \) vertically.3. The normal force \( \vec{N} \) acting horizontally.4. The kinetic friction force \( \vec{f}_k \) acting downwards along the wall.

Step 3: Resolve the applied force

Resolve the force \( \vec{F} \) into horizontal and vertical components:- Horizontal component: \( F_x = 60 \cos 60^{\circ} = 30 \text{ N} \)- Vertical component: \( F_y = 60 \sin 60^{\circ} = 51.96 \text{ N} \)

Step 4: Apply Newton's Second Law in the vertical direction

For constant velocity, the acceleration is zero, so net force vertically is zero:\( F_y - f_k - F_g = 0 \)Substitute the known values:\( 51.96 - f_k - 29.4 = 0 \)

Step 5: Apply Newton's Second Law in the horizontal direction

In the horizontal direction, since the object does not move into or away from the wall, the net force is zero:\( N - F_x = 0 \)So, \( N = F_x = 30 \text{ N} \)

Step 6: Solve for the force of kinetic friction

Solve the equation from Step 4 for \( f_k \):\( f_k = 51.96 - 29.4 = 22.56 \text{ N} \)

Step 7: Compile Results

Based on the Newton's Second Law calculations:- The normal force on the object is \( 30 \text{ N} \).- The force of kinetic friction is \( 22.56 \text{ N} \).

Key Concepts

Free Body Diagram

When analyzing a physics problem involving multiple forces, like an object moving up a wall, the free body diagram is an essential starting point. It visually represents all the forces acting on the object.
For the given exercise, let's break down the forces:

• Gravitational Force: This force acts downward due to gravity and is represented as \( \vec{F}_g \). For our 3.0 kg mass, this force is \( F_g = mg = 29.4 \text{ N}\) downward.
• Applied Force: This is the force causing motion. It's given as \( 60 \text{ N} \) at a \( 60^{\circ} \) angle to the horizontal. It has two components:
  • Horizontal Component: \( F_x = F \cos 60^{\circ} = 30 \text{ N} \)
  • Vertical Component: \( F_y = F \sin 60^{\circ} = 51.96 \text{ N} \)
• Normal Force: This force is exerted by the wall on the object, acting perpendicular to the wall.
• Kinetic Friction: Opposes the object's movement up the wall and acts downward along the wall surface.

These components are vectors and should be accurately represented in the diagram to ensure a complete analysis of the forces.

Kinetic Friction

Kinetic friction is the force that opposes the movement of two surfaces sliding past one another. It acts in the direction opposite to the motion.
For our scenario where a box slides upward on a wall:

• The kinetic friction, \( \vec{f}_k \), is acting downward, opposing the upward motion caused by the applied force.
• The magnitude of kinetic friction can be determined using Newton's second law. In this exercise, we found it to be \( 22.56 \text{ N} \).

This frictional force plays a crucial role as it affects the required applied force needed to maintain constant velocity. Often confused with static friction, note that kinetic friction applies only when there is motion. Recognizing its impact helps us understand why certain amounts of force are needed to slide objects along surfaces.

Normal Force

The normal force is a key component in understanding how objects interact with surfaces. It acts perpendicular to the surface to counterbalance the object's weight and any other forces pressing against the surface.
In the context of the object sliding up a wall:

• The normal force, \( \vec{N} \), acts horizontally from the wall back towards the object.
• Using the horizontal equation of motion \( N - F_x = 0 \), we determined that the normal force is equal to the horizontal component of the applied force, which is \( 30 \text{ N} \).

Understanding the normal force is vital as it influences the amount of frictional force present. It's not just counteracting gravity; it also adjusts based on forces applied, such as the horizontal component of the applied force in this exercise. This makes it a crucial factor in determining the motion and equilibrium of objects on surfaces.