VECTOR MECHANICS FOR ENGINEERS: STATICS
VECTOR MECHANICS FOR ENGINEERS: STATICS
12th Edition
ISBN: 9781260912814
Author: BEER
Publisher: MCG
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Chapter 8.1, Problem 8.13P

Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C, the coefficients of friction are μs = 0.30 and μk = 0.20; between package B and the belt, the coefficients are μs = 0.10 and μk = 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.

Chapter 8.1, Problem 8.13P, Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and

Fig. P8.13

Expert Solution & Answer
Check Mark
To determine

Find whether any of the package moves and the friction force acting on each package.

Answer to Problem 8.13P

The package C will notmove_.

The friction force in the package C is FC=10.16N()_.

The packages A and B will move_.

The friction force in the package B is FB=3.03N()_.

The friction force in the package A is FA=7.58N()_.

Explanation of Solution

Given information:

The mass of the package A, B, and C is mA=mB=mC=4kg.

The static coefficient of friction between packages A and C and the belt is

(μs)A=(μs)C=0.30.

The static coefficient of friction between package B and belt is (μs)B=0.10.

The kinetic coefficient of friction between packages A and C and belt is

(μk)A=(μk)C=0.20.

The kinetic coefficient of friction between package B and belt is (μk)B=0.08.

Calculation:

Consider the acceleration due to gravity as g=9.81m/s2.

Consider Block C:

Show the free body diagram of the block C as in Figure 1.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 8.1, Problem 8.13P , additional homework tip  1

Resolve the vertical component of forces.

+Fy=0NCmCgcos15°=0NC4×9.81cos15°=0NC=37.9N()

Resolve the horizontal component of forces.

Fx=0FCmCgsin15°=0FC4×9.81sin15°=0FC=10.16N()

Find the maximum friction force (Fm) using the relation.

(Fm)C=(μs)CNC

Substitute 0.30 for (μs)C and 37.9 N for NC.

(Fm)C=0.30×37.9=11.37N

The maximum friction force is greater than the friction force.

(Fm)C=11.37N>FC=10.16N

Therefore, the package C will notmove_.

Therefore, the friction force in the package C is FC=10.16N()_.

Consider Block B:

Show the free body diagram of the block B as in Figure 2.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 8.1, Problem 8.13P , additional homework tip  2

Resolve the vertical component of forces.

+Fy=0NBmBgcos15°=0NB4×9.81cos15°=0NB=37.9N()

Resolve the horizontal component of forces.

Fx=0FBmBgsin15°=0FB4×9.81sin15°=0FB=10.16N()

Find the maximum friction force (Fm)B using the relation.

(Fm)B=(μs)BNB

Substitute 0.10 for (μs)B and 37.9 N for NB.

(Fm)B=0.10×37.9=3.79N

The maximum friction force is less than the friction force.

(Fm)B=3.79N<FB=10.16N

Therefore, the package B will move_.

Find the friction force in the package B using the kinetic relation.

FB=(μk)BNB

Substitute 0.08 for (μk)B and 37.9 N for NB.

FB=0.08×37.9=3.03N

Therefore, the friction force in the package B is FB=3.03N()_.

Consider Block A and B together:

Show the free body diagram of the block A and B as in Figure 3.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 8.1, Problem 8.13P , additional homework tip  3

The normal force in package A is NA=NC=37.9N().

The normal force in package B is NB=37.9N().

The friction force in package A is FA=FC=10.16N().

The friction force in package B is FB=10.16N().

Find the total normal force in package A and B as follows;

NA+NB=37.9+37.9=75.8N()

Find the total friction force in package A and B as follows;

FA+FB=10.16+10.16=20.32N()

The maximum friction force in package A is (Fm)A=(Fm)C=11.37N.

The maximum friction force in package B is (Fm)B=3.79N.

Find the maximum friction force (Fm)A+B using the relation.

(Fm)A+B=(Fm)A+(Fm)B=11.37+3.79=15.16N

The maximum friction force is less than the friction force.

(Fm)A+B=15.16N<FA+FB=20.32N

Therefore, the packages A and B will move_.

Find the friction force in the package A using the kinetic relation.

FA=(μk)ANA

Substitute 0.20 for (μk)A and 37.9 N for NA.

FA=0.20×37.9=7.58N()

Therefore, the friction force in the package A is FA=7.58N()_.

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Chapter 8 Solutions

VECTOR MECHANICS FOR ENGINEERS: STATICS

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The...Ch. 8.2 - The spring of the door latch has a constant of 1.8...Ch. 8.2 - Prob. 8.61PCh. 8.2 - Prob. 8.62PCh. 8.2 - Prob. 8.63PCh. 8.2 - A 15 wedge is forced under a 50-kg pipe as shown....Ch. 8.2 - A 15 wedge is forced under a 50-kg pipe as shown....Ch. 8.2 - Prob. 8.66PCh. 8.2 - Prob. 8.67PCh. 8.2 - Derive the following formulas relating the load W...Ch. 8.2 - The square-threaded worm gear shown has a mean...Ch. 8.2 - Prob. 8.70PCh. 8.2 - High-strength bolts are used in the construction...Ch. 8.2 - The position of the automobile jack shown is...Ch. 8.2 - For the jack of Prob. 8.72, determine the...Ch. 8.2 - Prob. 8.74PCh. 8.2 - Prob. 8.75PCh. 8.2 - Prob. 8.76PCh. 8.3 - A lever of negligible weight is loosely fitted...Ch. 8.3 - A 6-in.-radius pulley of weight 5 lb is attached...Ch. 8.3 - 8.79 and 8.80 The double pulley shown is attached...Ch. 8.3 - Prob. 8.80PCh. 8.3 - 8.81 and 8.82 The double pulley shown is attached...Ch. 8.3 - 8.81 and 8.82 The double pulley shown is attached...Ch. 8.3 - The block and tackle shown are used to raise a...Ch. 8.3 - The block and tackle shown are used to lower a...Ch. 8.3 - A scooter is to be designed to roll down a 2...Ch. 8.3 - The link arrangement shown is frequently used in...Ch. 8.3 - 8.87 and 8.88 A lever AB of negligible weight is...Ch. 8.3 - 8.87 and 8.88 A lever AB of negligible weight is...Ch. 8.3 - 8.89 and 8.90 A lever AB of negligible weight is...Ch. 8.3 - 8.89 and 8.90 A lever AB of negligible weight is...Ch. 8.3 - A loaded railroad car has a mass of 30 Mg and is...Ch. 8.3 - Prob. 8.92PCh. 8.3 - A 50-lb electric floor polisher is operated on a...Ch. 8.3 - The frictional resistance of a thrust bearing...Ch. 8.3 - Assuming that bearings wear out as indicated in...Ch. 8.3 - Assuming that the pressure between the surfaces of...Ch. 8.3 - Solve Prob. 8.93 assuming that the normal force...Ch. 8.3 - Determine the horizontal force required to move a...Ch. 8.3 - Knowing that a 6-in.-diameter disk rolls at a...Ch. 8.3 - A 900-kg machine base is rolled along a concrete...Ch. 8.3 - Solve Prob. 8.85 including the effect of a...Ch. 8.3 - Solve Prob. 8.91 including the effect of a...Ch. 8.4 - A rope having a weight per unit length of 0.4...Ch. 8.4 - A hawser is wrapped two full turns around a...Ch. 8.4 - Two cylinders are connected by a rope that passes...Ch. 8.4 - Two cylinders are connected by a rope that passes...Ch. 8.4 - The coefficient of static friction between block B...Ch. 8.4 - The coefficient of static friction S is the same...Ch. 8.4 - A band belt is used to control the speed of a...Ch. 8.4 - The setup shown is used to measure the output of a...Ch. 8.4 - The setup shown is used to measure the output of a...Ch. 8.4 - A flat belt is used to transmit a couple from drum...Ch. 8.4 - A flat belt is used to transmit a couple from...Ch. 8.4 - Prob. 8.114PCh. 8.4 - The speed of the brake drum shown is controlled by...Ch. 8.4 - Prob. 8.116PCh. 8.4 - The speed of the brake drum shown is controlled by...Ch. 8.4 - Bucket A and block C are connected by a cable that...Ch. 8.4 - Solve Prob. 8.118 assuming that drum B is frozen...Ch. 8.4 - Prob. 8.120PCh. 8.4 - 8.121 and 8.123 A cable is placed around three...Ch. 8.4 - Prob. 8.122PCh. 8.4 - 8.121 and 8.123 A cable is placed around three...Ch. 8.4 - A recording tape passes over the 20-mm-radius...Ch. 8.4 - Solve Prob. 8.124 assuming that the idler drum C...Ch. 8.4 - Prob. 8.126PCh. 8.4 - The axle of the pulley is frozen and cannot rotate...Ch. 8.4 - Prob. 8.128PCh. 8.4 - Prob. 8.129PCh. 8.4 - Prove that Eqs. 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