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Chapter 6: Problem 11

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15掳 above the horizontal. (a) If the coefficient of static friction is 0.65, what minimum force magnitude is required from the rope to start the crate moving? (b) If mk 0.35, what is the magnitude of the initial acceleration of the crate?

Short Answer

Expert verified
(a) F = 541 N is required to start moving. (b) Acceleration when moving is 1.40 m/s虏.

Step by step solution

01

Identify Forces

We need to identify all the forces acting on the crate. The force we apply (\( F \)) has two components due to the angle: a horizontal component \( F \cos(15^\circ) \) and a vertical component \( F \sin(15^\circ) \). The normal force is affected by the vertical component of the applied force. The static friction force can be calculated as \( f_s = \mu_s N \) where \( N \) is the normal force.
02

Apply Equilibrium in Vertical Direction

In the vertical direction, the normal force (\( N \)) balances the weight and the vertical component of the pulling force. The equations in the vertical direction are given as: \[N + F \sin(15^\circ) = Mg \]where \( N \) is the normal force, \( M \) is the mass of the crate, and \( g \) is the acceleration due to gravity (9.8 m/s虏). Thus, \( N = Mg - F \sin(15^\circ) \).
03

Calculate Static Friction Force

The maximum static friction force before the crate starts moving is given by \( f_s = \mu_s N \). Thus:\[ f_s = 0.65 \times (Mg - F \sin(15^\circ)) \]This needs to equal the horizontal component of the pulling force \( F \cos(15^\circ) \) for the crate to start moving.
04

Solve for Minimum Force Required

Set the horizontal component of the force equal to the static friction force and solve for \( F \):\[ F \cos(15^\circ) = 0.65 (Mg - F \sin(15^\circ)) \]Plug in \( M = 68 \) kg, \( g = 9.8 \) m/s虏:\[ F \cos(15^\circ) = 0.65 (68 \times 9.8 - F \sin(15^\circ)) \]Calculate \( F \) to find the minimum magnitude required.
05

Calculate Net Force in Horizontal Direction for Part (b)

After the crate starts moving, the kinetic friction comes into play. The kinetic friction force is:\[ f_k = \mu_k N = 0.35 (Mg - F \sin(15^\circ)) \]Net force \( F_{net} \):\[ F_{net} = F \cos(15^\circ) - f_k \]
06

Calculate Acceleration

Using Newton's second law, the net force equals mass times acceleration:\[ F_{net} = Ma \]Rearrange to find the acceleration \( a \) and substitute known values:\[ a = \frac{F_{net}}{M} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Friction
Static friction is the force that keeps an object at rest when a force is applied until that force exceeds a certain threshold. Before a crate starts moving, static friction acts to resist the applied force. The amount of static friction is determined by multiplying the coefficient of static friction (\(\mu_s\)) by the normal force (\(N\)).
  • Coefficient of Static Friction: This is a dimensionless value that represents how much friction force is present between two surfaces at rest. For our problem, we have \(\mu_s = 0.65\).
  • Normal Force: The normal force is the perpendicular force exerted by a surface against an object in contact. It is calculated considering both the weight of the object and any vertical components of applied forces. For the crate, we calculated \(N = Mg - F\sin(15^\circ)\).
Once the applied horizontal force component equals the static friction, the crate will start moving.
Kinetic Friction
Once the crate begins to move, static friction is replaced by kinetic friction, which is normally less than static friction. Kinetic friction acts in the opposite direction of motion and affects how easily an object continues to move.
  • Coefficient of Kinetic Friction: Designated as \(\mu_k\), this coefficient is lower than that of static friction (0.35 in our example).
  • Kinetic Friction Force: It is computed by multiplying \(\mu_k\) by the normal force \(N\), so \(f_k = \mu_kN\).
Understanding the difference between static and kinetic friction is pivotal to predicting and calculating the forces and motion of objects as they transition from rest to motion.
Newton's Second Law
Newton's second law of motion is crucial for understanding forces and motion, and it states that the net force acting on an object is equal to the product of its mass and acceleration (\(F_{net} = Ma\)). In our problem, this principle helps derive the acceleration once the crate is in motion.
  • Net Force Calculation: Once the crate moves, the net force is difference between the horizontal pulling force and the kinetic friction force (\(F_{net} = F\cos(15^\circ) - f_k\)).
  • Solving for Acceleration: Rearrange the second law formula to find acceleration as \(a = \frac{F_{net}}{M}\), which provides insight into how quickly the crate accelerates across the floor.
By applying this law, it becomes easier to predict and calculate the behavior of objects under various forces, aiding in a comprehensive understanding of motion.
Forces and Motion
Forces are any interaction that, when unopposed, change the motion of an object. In this scenario, the applied force, normal force, gravitational force, static frictional force, and kinetic frictional force collectively determine the crate's motion.
  • Applied Force: A force exerted through a rope at 15掳 above the horizontal. This affects both the vertical and horizontal force components.
  • Motion Initiation: The horizontal component of the applied force must surpass the maximum static friction to start moving the crate.
  • Continuous Motion: As the crate starts moving, kinetic friction influences its continued motion, dictating the need for consistent net force to sustain acceleration.
By mastering forces and motion, students can predict outcomes in physics problems and understand real-world applications of these principles in everyday scenarios.

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Most popular questions from this chapter

A puck of mass m 1.50 kg slides in a circle of radius r 25.0 cm on a frictionless table while attached to a hanging cylinder of mass M 2.50 kg by means of a cord that extends through a hole in the table (Fig. 6-45). What speed keeps the cylinder at rest?

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.0 kN, and the circle鈥檚 radius is 10 m. At the top of the circle, what are the (a) magnitude FB and (b) direction (up or down) of the force on the car from the boom if the car鈥檚 speed is v 5.0 m/s? What are (c) FB and (d) the direction if v 12 m/s?

The mysterious sliding stones. Along the remote Racetrack Playa in Death Valley, California, stones sometimes gouge out prominent trails in the desert floor, as if the stones had been migrating (Fig. 6-16). For years curiosity mounted about why the stones moved. One explanation was that strong winds during occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force must act on a 20 kg stone (a typical mass) to maintain the stone鈥檚 motion once a gust has started it moving? (Story continues with Problem 37.)

A baseball player with mass m 83 kg, sliding into second base, is retarded by a frictional force of magnitude 485 N. What is the coefficient of kinetic friction mk between the player and the ground?

5 A student of weight 667 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force on the student from the seat is 556 N. (a) Does the student feel 鈥渓ight鈥� or 鈥渉eavy鈥� there? (b) What is the magnitude of at the lowest point? If the wheel鈥檚 speed is doubled, what is the magnitude FN at the (c) highest and (d) lowest point?
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