91Ó°ÊÓ

A bookcase weighing 1500 \(\mathrm{N}\) rests on a horizon- tal surface for which the coefficient of static friction is \(\mu_{\mathrm{s}}=0.40 .\) The bookcase is 1.80 \(\mathrm{m}\) tall and 2.00 \(\mathrm{m}\) wide; its center of gravity is at its geo- metrical center. The bookcase rests on four short legs that are each 0.10 \(\mathrm{m}\) from the edge of the bookcase. A person pulls on a rope attached to an upper corner of the bookcase with a force \(\vec{\boldsymbol{F}}\) that makes an angle \(\theta\) with the bookcase (Fig. \(P 11.97 )\) . (a) If \(\theta=90^{\circ},\) so \(\vec{F}\) is horizontal, show that as \(F\) is increased from zero, the bookcase will start to slide before it tips, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will start the bookcase sliding. (b) If \(\theta=0^{\circ},\) so \(\vec{\boldsymbol{F}}\) is vertical, show that the bookcase will tip over rather than slide, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to tip. (c) Calculate as a function of \(\theta\) the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to slide and the magnitude that will cause it to start to tip. What is the smallest value that \(\theta\) can have so that the bookcase will still start to slide before it starts to tip?

Step by step solution

01

Determine the Static Friction Force

The static friction force can be calculated using the formula: \( f_s = \mu_s \cdot N \), where \( \mu_s = 0.40 \) is the coefficient of static friction and \( N = 1500 \, \text{N} \) is the normal force (equal to the weight of the bookcase). Therefore, \( f_s = 0.40 \times 1500 \, \text{N} = 600 \, \text{N} \).

02

Calculate Force to Slide (Horizontal Force)

For the bookcase to start sliding horizontally, the applied force must exceed the static friction force. Thus, the force \( F \) required to start sliding is 600 N when \( \theta = 90^{\circ} \).

03

Determine the Tipping Condition (Vertical Force)

For tipping at \( \theta = 0^{\circ} \): Calculate the torque about the pivot point (one of the front legs). The torque due to gravity is \( 1500 \, \text{N} \times 0.90 \, \text{m} \) (since the center of gravity is at the middle), and the torque due to the applied force \( F \) is \( 2.00 \, \text{m} \times F \). Setting these equal to find the tipping point: \( F_{\_\text{tip}} = \frac{1500 \, \text{N} \times 0.90 \, \text{m}}{2.00 \, \text{m}} = 675 \, \text{N} \).

04

Function for Sliding and Tipping Forces at Angle \(\theta\)

The force \( F \) causing sliding is influenced by the horizontal component only: \( F_{\_\text{slide}} = \frac{600}{\cos{\theta}} \). The force causing tipping is influenced by the vertical component: \( F_{\_\text{tip}} = \frac{1500 \, \text{N} \times 0.90 \, \text{m}}{2.00 \, \text{m} \times \sin{\theta}} = \frac{675}{\sin{\theta}} \).

05

Determine the Mixture Angle \(\theta\) for Sliding Before Tipping

Set \( F_{\_\text{slide}} = F_{\_\text{tip}} \) to find the angle where sliding and tipping forces equate: \( \frac{600}{\cos{\theta}} = \frac{675}{\sin{\theta}} \). Solve for \( \theta \): \( \tan{\theta} = \frac{675}{600} = \frac{9}{8} \). The smallest \( \theta \) for sliding before tipping is \( \theta = \arctan{\left(\frac{9}{8}\right)} \approx 48.37^{\circ} \).

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 91Ó°ÊÓ!

Key Concepts

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

Coefficient of Friction

The coefficient of friction is a measure of how much frictional force exists between two surfaces. It is represented by the symbol \( \mu \) and varies based on the materials in contact. In this exercise, we're dealing with static friction, which is the force that prevents an object from moving on a surface.
Static friction applies when an object is still at rest.

• The coefficient of static friction, \( \mu_s \), is 0.40 here, meaning it's quite moderate.
• The static friction force can be found using the formula: \( f_s = \mu_s \cdot N \).
• This means you multiply the coefficient by the normal force (which is the weight of the object if on a flat surface and stable).

To apply the learned concept: In this exercise, the total static friction force is calculated to be 600 N, serving as the threshold for the horizontal force needed to overcome to start the sliding process. A greater force is required to initiate movement owing to static friction than to maintain it due to kinetic friction.

Torque Calculation

Torque is the measure of the rotational force at play around a pivot point. It depends on the force applied and the distance from the pivot at which the force is applied. Here's how it applies:
Torque can cause objects to rotate or tip rather than slide. Torque is significant when forces act at angles or are applied at different heights.

• To calculate torque, use the formula: \( \tau = r \times F \), where \( r \) is the lever arm distance, and \( F \) is the perpendicular force.

In this example, we consider the pivot at the tip of one of the bookcase's legs:
- Calculate the gravitational torque that attempts to keep the bookcase stable, which is \( 1500 \, \text{N} \times 0.90 \, \text{m} \).
- Calculate the torque due to the vertical force \( F \) that's applied by the person.
li>The idea is to find when these torques become equal. This occurs at a force of 675 N, causing the bookcase to tip.

Forces and Angles

Forces and angles determine how an object reacts to applied forces. They influence whether it will slide off or tip over.
When applying a force, the angle of the force affects how it is divided into horizontal and vertical components.

• A horizontal force (\( \theta = 90^{\circ} \)) mainly counters friction.
• Whereas a vertical force (\( \theta = 0^{\circ} \)) influences tipping through torque.

In practice, an object's reaction depends on these interactions.
If force is applied at different angles, knowing the angle helps calculate the exact force needed using trigonometric functions. In this exercise, forces are divided using \( \cos \theta \) for horizontal components and \( \sin \theta \) for vertical ones.
To prevent it from tipping first, it's calculated that the smallest angle \( \theta \approx 48.37^{\circ} \) ensures sliding happens before tipping. Calculating this is crucial for effectively understanding the dynamics and predicting which event – sliding or tipping – will occur first.