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Physics Principles with Applications
Found in: Page 230
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

A 2.0-m-high box with a 1.0-m-square base is moved across a rough floor as in Fig. 9–89. The uniform box weighs 250 N and has a coefficient of static friction with the floor of 0.60. What minimum force must be exerted on the box to make it slide? What is the maximum height h above the floor that this force can be applied without tipping the box over? Note that as the box tips, the normal force and the friction force will act at the lowest corner.

  1. The minimum force that must be exerted on the box to make it slide is 150 N.
  2. The height h above the floor that must be piled on the box is 0.83 m.
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Identification of the given data

The length of the box is \(l = 2\;{\rm{m}}\).

The width of the box is \(b = 1\;{\rm{m}}\).

The weight of the box is \(W = 250\;{\rm{N}}\).

The coefficient of static friction between the box and the floor is \({\mu _{\rm{s}}} = 0.6\).

Step 2: Drawing the free body diagram (FBD) of the given situation

A free body diagram (FBD) is a simplified representation of a rigid body to visualize all the force vectors acting on it. The free body shown below depicts the forces on the uniform box.

The forces acting on the box are the following:

  • Weight of the box engine, \(W = mg\) vertically downwards
  • Normal force, N perpendicular to the floor
  • Applied force (pushing), F
  • Frictional force, \({F_{{\rm{fr}}}}\) between the floor and the box

Step 4: Analysis of the given situation

The uniform box is considered to be on the verge of both sliding and slipping yet in a state of equilibrium. The static friction will have a maximum value because the box is just about to slide.

Suppose the box is just about to slide and touch the floor, such that all the forces on the box due to the contact with the ground are at the lower right-hand corner.

Step 5: Evaluation of the forces parallel and perpendicular to the edge of the box

The forces along the y-direction (parallel to the edge of the box) are given by the equilibrium equation.

\(\begin{array}{c}\sum {{F_{\rm{y}}}} = 0\\{F_{\rm{N}}} - W = 0\\{F_{\rm{N}}} = W\end{array}\) … (i)

This shows that the normal force balances the weight (mg) of the box.

The forces along the x-axis (perpendicular to the edge of the box) are given as follows:

\(\begin{array}{c}\sum {{F_{\rm{x}}}} = 0\\F - {F_{{\rm{fr}}}} = 0\\F = {F_{{\rm{fr}}}}\\F = {\mu _{\rm{S}}}{F_{\rm{N}}}\end{array}\) … (ii)

Step 6: (a) Evaluation of the force that must be exerted on the box to make it slide

From equations (i) and (ii),

\(\begin{array}{c}F = {\mu _{\rm{S}}}W\\ = \left( {0.6} \right)\left( {250\;{\rm{N}}} \right)\\ = 150\;{\rm{N}}\end{array}\)

Step 7: (b) Evaluation of the maximum height h that must be applied to the box

At the critical moment when the box is just about to tip and slide, the net torque is equal to zero. Take the torques in the clockwise direction as positive.

Using the equation of equilibrium,

\(\begin{array}{r}\sum {\tau = 0} \\Fh - W\left( {\frac{b}{2}} \right) = 0\\h = \frac{W}{F}\left( {\frac{b}{2}} \right)\end{array}\)

Substituting numerical values in the above expression gives the height h above the floor, where the force should be applied without tipping the box.

\(\begin{array}{c}h = \frac{{250\;{\rm{N}}}}{{150\;{\rm{N}}}}\left( {\frac{{1\;{\rm{m}}}}{2}} \right)\\ = 0.83\;{\rm{m}}\end{array}\)

Most popular questions for Physics Textbooks

When a wood shelf of mass 6.6 kg is fastened inside a slot in a vertical support as shown in Fig. 9–92, the support exerts a torque on the shelf. (a) Draw a free-body diagram for the shelf, assuming three vertical forces (two exerted by the support slot—explain why). Then calculate (b) the magnitudes of the three forces and (c) the torque exerted by the support (about the left end of the shelf).

A parking garage is designed for two levels of cars. To make more money, the owner decides to double the size of the garage in each dimension (length, width, and the number of levels). For the support columns to hold up four floors instead of two, how should he change the columns' diameter?

(a) Double the area of the columns by increasing their diameters by a factor of 2

(b) Double the area of the columns by increasing their diameters by a factor of \(\sqrt 2 \)

(c) Quadruple the area of the columns by increasing their diameters by a factor of 2

(d) Increase the area of the columns by a factor of 8 by increasing their diameters by a factor of \(2\sqrt 2 \)

(e) He doesn't need to increase the diameter of the columns

A 23.0-kg backpack is suspended midway between two trees by a light cord as in Fig. 9–51. A bear grabs the backpack and pulls vertically downward with a constant force, so that each section of cord makes an angle of 27° below the horizontal. Initially, without the bear pulling, the angle was 15°; the tension in the cord with the bear pulling is double what it was when he was not. Calculate the force the bear is exerting on the backpack.

(II) The subterranean tension ring that exerts the balancing horizontal force on the abutments for the dome in Fig. 9–34 is 36-sided, so each segment makes a 10° angle with the adjacent one (Fig. 9–77). Calculate the tension F that must exist in each segment so that the required force of \(4.2 \times {10^5}\;{\rm{N}}\) can be exerted at each corner (Example 9–13).

A 15.0-kg ball is supported from the ceiling by rope A. Rope B pulls downward and to the side on the ball. If the angle of A to the vertical is 22° and if B makes an angle of 53° to the vertical (Fig. 9–85), find the tensions in ropes A and B.
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