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8 1 The mine car and its contents have a total mass of 6 Mg and a center of gravity at G If the coefficient of static friction between the wheels and the tracks is ms 0 4 when the wheels are locked find the normal force acting on the front wheels at B and the rear wheels at A when the brakes at both A and B are locked Does the car move 10 kN 0 9 m G B SOLUTION 1 5 m NA 11 52 1011 052 58 8610 62 0 NA 16 544 kN 16 5 kN c Fy 0 0 15 m 0 6 m Equations of Equilibrium The normal reactions acting on the wheels at A and B are independent as to whether the wheels are locked or not Hence the normal reactions acting on the wheels are the same for both cases a MB 0 A Ans NB 16 544 58 86 0 Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b NB 42 316 kN 42 3 kN When both wheels at A and B are locked then 1FA2max msNA 0 4116 5442 6 6176 kN and 1FB2max msNB 0 4142 3162 16 9264 kN Since 1FA2max FB max 23 544 kN 7 10 kN the wheels do not slip Thus the mine car does not move Ans 2013 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permission s write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 8 2 P 2 P 2 Determine the maximum force P the connection can support so that no slipping occurs between the plates There are four bolts used for the connection and each is tightened so that it is subjected to a tension of 4 kN The coefficient of static friction between the plates is ms 0 4 SOLUTION Free Body Diagram The normal reaction acting on the contacting surface is equal to the sum total tension of the bolts Thus N 4 4 kN 16 kN When the plate is on the verge of slipping the magnitude of the friction force acting on each contact surface can be computed using the friction formula F msN 0 4 16 kN As indicated on the free body diagram of the upper plate F acts to the right since the plate has a tendency to move to the left Equations of Equilibrium 0 4 16 P 0 2 p 12 8 kN Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b Fx 0 2013 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permission s write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 P 8 3 The winch on the truck is used to hoist the garbage bin onto the bed of the truck If the loaded bin has a weight of 8500 lb and center of gravity at G determine the force in the cable needed to begin the lift The coefficients of static friction at A and B are mA 0 3 and mB 0 2 respectively Neglect the height of the support at A 30 G A 10 ft 12 ft B SOLUTION a MB 0 8500 12 NA 22 0 NA 4636 364 lb F 0 x T cos 30 0 2NB cos 30 NB sin 30 0 3 4636 364 0 T 0 86603 0 67321 NB 1390 91 4636 364 8500 T sin 30 NB cos 30 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b c Fy 0 0 2NB sin 30 0 T 0 5 0 766025 NB 3863 636 Solving T 3666 5 lb 3 67 kip Ans NB 2650 6 lb 2013 Pearson Education Inc Upper Saddle River NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction storage in a retrieval system or transmission in any form or by any means electronic mechanical photocopying recording or likewise For information regarding permission s write to Rights and Permissions Department Pearson Education Inc Upper Saddle River NJ 07458 8 4 The tractor has a weight of 4500 lb with center …
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