/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A worker pushes horizontally on ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 13

A worker pushes horizontally on a 35 kg crate with a force of magnitude 110 N.The coefficient of static friction between the crate and the floor is 0.37. (a) What is the value of fs,max under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker鈥檚 110 N push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

Short Answer

Expert verified
(a) 126.91 N; (b) No; (c) 110 N; (d) 45 N; (e) 16.91 N.

Step by step solution

01

Calculate the maximum static friction (fs,max)

The maximum static friction is given by the formula \( f_{s,\text{max}} = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. The normal force in this scenario is the weight of the crate, which is calculated as \( N = m \cdot g = 35 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 343 \text{ N} \). Thus, \( f_{s,\text{max}} = 0.37 \cdot 343 \text{ N} = 126.91 \text{ N} \).
02

Determine if the crate moves

We compare the applied horizontal force of 110 N with the maximum static friction \( f_{s,\text{max}} = 126.91 \text{ N} \). Since the applied force is less than the maximum static friction, the crate does not move.
03

Evaluate the frictional force on the crate

Since the crate does not move, the static frictional force \( f_s \) exactly balances the applied force, which is 110 N. The frictional force is thus 110 N.
04

Calculate the least vertical pull to move the crate

To move the crate, the vertical pull \( F_v \) will reduce the normal force \( N \), making the normal force \( N' = 343 \text{ N} - F_v \). We need \( 110 \text{ N} \geq 0.37 \cdot N' \) to hold. Solving for \( F_v \) gives us: \[ 110 \text{ N} = 0.37 \times (343 \text{ N} - F_v) \] \[ F_v = 343 \text{ N} - \frac{110}{0.37} \approx 45 \text{ N} \].
05

Calculate the least horizontal pull to move the crate

With the second worker also pushing horizontally, the total horizontal force required to overcome static friction is still \( f_{s,\text{max}} = 126.91 \text{ N} \). Since one worker already applies 110 N, the least pull needed by the second worker is therefore \( 126.91 \text{ N} - 110 \text{ N} = 16.91 \text{ N} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Coefficient of Static Friction
The coefficient of static friction, denoted as \( \mu_s \), is a measure of how much frictional force exists between two still surfaces. In this exercise, it quantifies how much force is required to start moving the crate from rest. The larger the coefficient, the more "sticky" the surfaces are to each other. It essentially represents how much the surfaces "grip" one another.
  • This value is dimensionless, meaning it has no units and is just a ratio.
  • In the given problem, \( \mu_s = 0.37 \), indicating a moderate level of grip between the crate and the floor.
Understanding \( \mu_s \) helps in predicting whether a certain amount of applied force will overcome the friction enough to move the object. It's a crucial parameter in scenarios involving stationary objects needing a push to start moving.
Normal Force
The normal force \( N \) is the force exerted by a surface that supports the weight of an object resting on it. It acts perpendicular to the surface. In this problem, the crate's weight generates the normal force, given by the equation \( N = m \cdot g \), where \( m \) is the mass of the crate, and \( g \) is the gravitational acceleration, \( 9.8 \, \text{m/s}^2 \).
  • For this crate, \( N = 35 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 343 \, \text{N} \).
  • This normal force is crucial since it directly affects the maximum static friction \( f_{s,\text{max}} \), calculated via \( \mu_s \cdot N \).
By understanding the normal force, we comprehend the total force available to resist other forces, like static friction in this scenario.
Frictional Force
Frictional force is the resistive force arising when two surfaces are in contact, preventing or slowing motion. In this context, static friction acts to prevent the movement of the crate until a certain force threshold is reached. This threshold is known as the maximum static friction \( f_{s,\text{max}} \).
  • In our example, \( f_{s,\text{max}} = 126.91 \, \text{N} \).
  • The frictional force aligns with the applied force of 110 N since the crate doesn't move. This means the static frictional force is also 110 N.
Understanding frictional force is crucial because it determines the amount of force needed to initiate the movement of stationary objects. If an applied force is less than \( f_{s,\text{max}} \), the object remains static.
Net Force Calculation
Net force calculation involves evaluating all the forces acting on an object to determine its motion. It is computed by summing up force vectors acting on the object. The net force's direction and magnitude affect how and whether an object moves.
  • If the net force equals zero, like in the initial situation with this crate, the object remains stationary.
  • To move the crate, the net force acting on it must be positive in the direction of intended motion, overcoming static friction.
To find the net force, one sums all horizontal forces and all vertical forces separately. Calculating the net force helps decide if an object will start moving, accelerate in a certain direction, or remain still. In scenarios where additional help (like a second worker) is added, these calculations adjust to find the minimum additional force needed to move the object.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 roller-coaster car has a mass of 1300 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 20 m, its speed is not changing. At the top of the hill, what are the (a) magnitude FN and (b) direction (up or down) of the normal force on the car from the track if the car鈥檚 speed is v 11 m/s? What are (c) FN and (d) the direction if v 14 m/s?

Calculate the ratio of the drag force on a jet flying at 1200 km/h at an altitude of 15 km to the drag force on a prop-driven transport flying at half that speed and altitude. The density of air is 0.38 kg/m3 at 10 km and 0.67 kg/m3 at 5.0 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C.

During an Olympic bobsled run, the Jamaican team makes a turn of radius 7.6 m at a speed of 96.6 km/h. What is their acceleration in terms of g?

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?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Translational Dynamics

Read Explanation

Fluids

Read Explanation

Space Physics

Read Explanation

Quantum Physics

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.