Problem 40
Question
A person pushes on a stationary \(125 \mathrm{~N}\) box with \(75 \mathrm{~N}\) at \(30^{\circ}\) below the horizontal, as shown in Figure 5.61 . The coefficient of static friction between the box and the horizontal floor is 0.80 . (a) Make a free-body diagram of the box. (b) What is the normal force on the box? (c) What is the friction force on the box? (d) What is the largest the friction force could be? (e) The person now replaces his push with a \(75 \mathrm{~N}\) pull at \(30^{\circ}\) above the horizontal. Find the normal force on the box in this case.
Step-by-Step Solution
Verified Answer
(a) Draw forces: weight, normal, friction, and applied forces.
(b) Normal force is 162.5 N.
(c) Friction force is 130 N.
(d) Maximum friction force is 130 N.
(e) Normal force with pull is 87.5 N.
1Step 1: Draw Free-Body Diagram
Identify all forces acting on the box. These include the gravitational force (weight) acting directly downward, the normal force acting upward, the static friction force acting horizontally opposite to any movement, and the applied force at an angle of 30° below the horizontal.
2Step 2: Determine the Normal Force with Push
The normal force can be found using the vertical force balance. Let the gravitational force be \( F_g \). Analyzing vertical components, we have: \( F_n = F_g + F_{push} \sin(30^{\circ}) \). Substitute \( F_g = 125 \text{ N} \) and \( F_{push} = 75 \text{ N} \) to get \( F_n = 125 + 75 \sin(30^{\circ}) = 162.5 \text{ N} \).
3Step 3: Calculate the Static Friction Force
Static friction force is given by \( f_s = \mu_s F_n \), where \( \mu_s = 0.80 \) is the coefficient of static friction. Using the calculated normal force from Step 2, \( f_s = 0.80 \times 162.5 = 130 \text{ N} \).
4Step 4: Determine Maximum Static Friction Force
Maximum static friction force is also \( f_{s_{max}} = \mu_s F_n \). However, since it occurs just before the box begins to move, it is the same value as calculated previously, so \( f_{s_{max}} = 130 \text{ N} \).
5Step 5: Find Normal Force with Pull
When the person pulls, the vertical component of the force acts upwards. Thus, the normal force becomes \( F_n = F_g - F_{pull} \sin(30^{\circ}) \). Substituting the values gives \( F_n = 125 - 75 \sin(30^{\circ}) = 87.5 \text{ N} \).
Key Concepts
Normal ForceFree-Body DiagramFriction ForcePhysics Problem Solving
Normal Force
Normal force is an essential concept in physics, especially when dealing with surfaces in contact. This force acts perpendicular to the surface of contact, often counteracting the force of gravity. In our problem, the normal force on the box plays a crucial role in determining the friction force between the box and the floor.
When a person pushes the box downwards at an angle, the normal force becomes larger. This happens because part of the downward force by the person (the component acting vertically) adds to the weight of the box. Specifically, in our problem, the gravitational force (\( F_g \)), which is 125 N, combines with the vertical component of the applied force. Using simple trigonometry, we find that component to be \( 75 \sin(30^{\circ}) \), which adds up to make the normal force \( 162.5 \, \text{N} \).
On the other hand, when the box is pulled upwards, the normal force decreases. Here, the vertical component of the pull subtracts from the weight of the box, making the normal force \( 87.5 \, \text{N} \). This change affects the friction force as it depends directly on the normal force.
When a person pushes the box downwards at an angle, the normal force becomes larger. This happens because part of the downward force by the person (the component acting vertically) adds to the weight of the box. Specifically, in our problem, the gravitational force (\( F_g \)), which is 125 N, combines with the vertical component of the applied force. Using simple trigonometry, we find that component to be \( 75 \sin(30^{\circ}) \), which adds up to make the normal force \( 162.5 \, \text{N} \).
On the other hand, when the box is pulled upwards, the normal force decreases. Here, the vertical component of the pull subtracts from the weight of the box, making the normal force \( 87.5 \, \text{N} \). This change affects the friction force as it depends directly on the normal force.
Free-Body Diagram
A free-body diagram is a simplified representation of an object and all the forces acting on it. This visual tool is fundamental in physics problem-solving, allowing students to isolate and understand the interactions and forces at play.
In drawing the free-body diagram for the box scenario described, you consider:
In drawing the free-body diagram for the box scenario described, you consider:
- The gravitational force (or the box’s weight) acting downward, represented by an arrow pointing down towards the ground.
- The normal force pointing upwards, balancing part of the downward forces.
- The applied force by the person, at an angle of \(30^{\circ}\), with components in both the horizontal and vertical directions.
- Finally, the static friction force, which opposes any potential movement, pointing horizontally opposite to the direction of the applied force.
Friction Force
Static friction force prevents the box from moving despite the pushing or pulling force applied to it. It acts parallel to the contact surface, opposing any motion. The magnitude of static friction depends on two factors: the coefficient of static friction \( (\mu_s) \) and the normal force \((F_n)\).
In our example, the static friction force can be found using the formula:\[ f_s = \mu_s \times F_n \]where \( \mu_s = 0.80 \). When calculated with the normal force from the pushing scenario, the static friction force is \( 130 \, \text{N} \).
This value changes with the force orientation. With a pulling force, the normal force decreases significantly because the vertical component of the applied force helps counteract the gravitational pull, resulting in diminished friction.
Understanding friction involves recognizing when the applied force overcomes static friction, leading to motion. This comprehension helps in determining whether or not the box will move, keeping real-world applications like tire-road interactions or furniture placement in mind.
In our example, the static friction force can be found using the formula:\[ f_s = \mu_s \times F_n \]where \( \mu_s = 0.80 \). When calculated with the normal force from the pushing scenario, the static friction force is \( 130 \, \text{N} \).
This value changes with the force orientation. With a pulling force, the normal force decreases significantly because the vertical component of the applied force helps counteract the gravitational pull, resulting in diminished friction.
Understanding friction involves recognizing when the applied force overcomes static friction, leading to motion. This comprehension helps in determining whether or not the box will move, keeping real-world applications like tire-road interactions or furniture placement in mind.
Physics Problem Solving
Problem-solving in physics can initially seem challenging. However, it becomes approachable with the right steps and understanding of core concepts like forces and motion.
1. Start by identifying all forces acting on an object. Utilize a free-body diagram to map these forces visually.
2. Apply the relevant equations based on the interactions observed in your diagram. Here, trigonometry helps decompose angled forces into horizontal and vertical components.
3. Check for force balances, particularly in cases of equilibrium where dynamic or static settings are involved. For instance, static equilibrium requires that all forces in each direction add up to zero.
4. Keep in mind different scenarios like pushing or pulling because they alter forces differently. As seen, an upward pull reduces the normal force compared to a downward push.
By practicing these steps consistently, you build a stronger intuition for tackling various physics problems, efficiently applying concepts such as friction and normal force dynamics in realistic scenarios.
1. Start by identifying all forces acting on an object. Utilize a free-body diagram to map these forces visually.
2. Apply the relevant equations based on the interactions observed in your diagram. Here, trigonometry helps decompose angled forces into horizontal and vertical components.
3. Check for force balances, particularly in cases of equilibrium where dynamic or static settings are involved. For instance, static equilibrium requires that all forces in each direction add up to zero.
4. Keep in mind different scenarios like pushing or pulling because they alter forces differently. As seen, an upward pull reduces the normal force compared to a downward push.
By practicing these steps consistently, you build a stronger intuition for tackling various physics problems, efficiently applying concepts such as friction and normal force dynamics in realistic scenarios.
Other exercises in this chapter
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