Problem 83 - College Physics

Question

\(\bullet$$\bullet\) You push with a horl- zontal fonce of 50 \(\mathrm{N}\) against a 20 \(\mathrm{N}\) box, press- ing it against a rough ver- tical wall to hold it in place, The coefficients of kinetic and static friction between this box and the wall are 0.20 and 0.50 , respectively. (a) Make a free-body diagram of this box. (b) What is the friction force on the box? (c) How hard would you have to press for the box to slide downward with a uni- form speed of 10.5 \(\mathrm{cm} / \mathrm{s}\) ?

Step-by-Step Solution

Answer

(a) Forces: weight down, normal force right, static friction up. (b) Friction force is 20 N upward. (c) Already slides at 10.5 cm/s with current setup; verify friction.

Step 1: Analyze the Forces and Draw Free-Body Diagram

**Free-Body Diagram**: Consider the box being pushed against a vertical wall with a horizontal force \( F_{h} = 50 \mathrm{N} \).- **Weight \( W \)**: The gravitational force acting downward. For a box with a weight \( W = 20 \mathrm{N} \).- **Normal Force \( N \)**: Exerted by the wall on the box, horizontally opposite to \( F_{h} \).- **Static Friction Force \( f_s \)**: Acts upward (if box is not sliding).Given forces:- Horizontal: \( F_{h} = N \), so the normal force \( N = 50 \mathrm{N} \).- Vertical: Static friction balances the weight \( W \) to prevent sliding.

Step 2: Calculate Static Friction Force

Use the coefficient of static friction \( \mu_s = 0.50 \).Static friction maximum force is given by:\[F_s = \mu_{s} \cdot N = 0.50 \times 50 \mathrm{N} = 25 \mathrm{N}\]Since the weight of the box \( W = 20 \mathrm{N} \) is less than \( F_s \), the box will not slide and the static friction force is equal to the weight.Thus, the static friction force \( f_s = 20 \mathrm{N} \) upwards to keep the box stationary.

Step 3: Calculate Force to Maintain Sliding

**Condition for Sliding**: Now, we want the box to slide downward at a constant speed. This uses kinetic friction:\( \mu_k = 0.20 \)The kinetic friction force \( f_k \) when sliding is:\[F_k = \mu_k \cdot N = 0.20 \times N = 0.20 \times 50 \mathrm{N} = 10 \mathrm{N}\]To slide down with constant speed, friction matches the component of gravity on the box:Therefore, when the frictional force \( f_k = 10 \mathrm{N} \), additional force required is to overcome the difference:Since gravity \( W = 20 \mathrm{N} \), the box will naturally slide. However, for constant speed, ensure the friction force is accounted:Thus, this part verifies \( f_k < W \); ensure enough horizontal force to have a kinetic friction only match weight.

Key Concepts

Free-body diagram

A free-body diagram is a useful tool in physics to visualize the forces acting on an object. In this exercise, we consider a box pressed against a vertical wall. The diagram includes:

• Horizontal force (\( F_h = 50 \mathrm{N} \)) exerting pressure on the box.
• Weight (\( W = 20 \mathrm{N} \)) acting downward due to gravity.
• Normal force (\( N \)), exerted by the wall, acting opposite to the applied horizontal force.
• Static friction force (\( f_s \)) acting upward when the box is stationary to prevent it from sliding.

The free-body diagram helps to balance the forces horizontally and vertically, ensuring that the box remains stationary until additional force causes motion.

Friction forces

Friction is what keeps objects from sliding or slipping by providing resistance against movement. It acts between surfaces in contact and can be understood as:

• Static friction: This force prevents the box from sliding when it's at rest. It balances the downward weight to keep the box stable against the wall.
• Kinetic friction: This force acts when the box begins to slide. It is generally less than static friction.

When the box is stationary against the wall, static friction comes into play by opposing the pull of gravity. As soon as the box moves, kinetic friction takes over, opposing the motion of the box sliding down the wall.

Kinetic and static friction

Static and kinetic friction are two forms of friction that act differently depending on whether an object is stationary or in motion.Static Friction:

• This occurs before the object starts to move.
• In the exercise, the box weighs 20 N, and static friction needs to balance this weight to keep it in place.
• Maximum static friction force is given by \( F_s = \mu_s \cdot N \).
• Here, \( \mu_s = 0.50 \), so the maximum static friction is \( 25 \mathrm{N} \).

Kinetic Friction:

• Takes over when the box slides.
• The coefficient is \( \mu_k = 0.20 \), leading to a kinetic friction force calculated as \( 10 \mathrm{N} \).
• Kinetic friction helps control the motion, ensuring the box slides at a constant speed.

Newton's laws of motion

Newton's laws of motion are fundamental when analyzing forces like in this exercise.

• First Law (Law of Inertia): A stationary box remains so unless acted upon by an unbalanced force. Here, static friction holds it in place as per this law.
• Second Law (F = ma): The law is used to calculate forces, ensuring balance in both directions. If the box has to slide, calculate the required force considering kinetic friction and weight.
• Third Law (Action-Reaction): The horizontal force applied on the box (\( 50 \mathrm{N} \)) is met with an equal and opposite normal force from the wall.

These laws help understand how different forces interact with each other and predict motion from given forces.