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Solutions to the third problem set....

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Problem Set #3

BME 110A

1. . Two children are standing on a diving board that weighs 146 lb. Knowing that the weights of the children at C and D are 63 lb and 90 lb, respectively, determine (a) the reaction at A, (b) the reaction at B.

Free-Body Diagram:

(a)

MB  0:

 Ay  3.6 ft  146 lb1.44 ft   63 lb 3.24 ft   90 lb 6.24 ft  0 or A y  271 lb

Ay   271.10 lb (b)

M A  0:

By 3.6 ft  146 lb 5.04 ft  63 lb 6.84 ft  90 lb9.84 ft 0 By  570.10 lb

or B y  570 lb

Page 1 of 7

Problem Set #3

BME 110A

2. The boom on a 9500-lb truck is used to unload a pallet of shingles of weight 3500 lb, Determine the reaction at each of the two (a ) rear wheels B, (b) front wheels C.

Free-Body Diagram:

(a) MC  0:

3.5 kips  1.6  1.3  19.5cos15  ft   2 FB 1.6  1.3  14 ft    9.5 kips1.6 ft  0 2FB  5.4009 kips

or FB  2.70 kips (b) M B  0:

 3.5 kips 19.5cos15



 14 ft   9.5 kips  14  1.3 ft   2 FC 14 1.3 1.6 ft   0 

2 FC  7.5991 kips, or

or FC  3.80 kips

Page 2 of 7

BME 110A

Problem Set #3

A hand truck is used to move two barrels, each weighing 80 lb. Neglecting the weight of the hand truck, determine (a) the vertical force P which should be applied to the handle to maintain equilibrium when   35 , (b) the corresponding reaction at each of the two wheels.

Free-Body Diagram:

a1   20 in. sin   8 in. cos a2  32 in. cos   20 in. sin  b  64 in. cos  From free-body diagram of hand truck M B  0: P  b   W  a2   W  a1   0

(1)

Fy  0: P  2w  2B  0

(2)

  35 

For

a1  20sin35  8cos35  4.9183 in. a2  32cos35  20sin35  14.7413 in.

b  64cos35  52.426 in. (a)

From Equation (1)

P 52.426 in.  80 lb 14.7413 in.  80 lb 4.9183 in.   0  P  14.9896 lb (b)

or P  14.99 lb 

From Equation (2)

14.9896 lb  2 80 lb   2B  0  B  72.505 lb

Page 3 of 7

or

B  72.5 lb 

BME 110A

Problem Set #3

4. In a laboratory experiment, students hang the weights shown from a beam of negligible weight. (a) Determine the reaction at the fixed support A knowing that end D of the beam does not touch support E. (b) Determine the reaction at the fixed support A knowing that the adjustable support E exerts an upward force of 1.2 lb on the beam.

Page 4 of 7

Problem Set #3

BME 110A (a)

Free-Body Diagram:

Fx  0:

Ax  0

Fy  0:

Ay  2 lb  1lb  0 Ay  3 lb or A  3 lb

M A  0:

MA – (2 lb)(8 in.) – (1 lb)(12 in.)  0 M A  28 lb  in.

(b)

or M A  28 lb  in.

Free Body Diagram:

Fx  0:

Fy  0:

Ax  0

Ay  2 lb  1 lb  1.2 lb  0 Ay  1.8 lb or A  1.8 lb

M A  0:

MA – (2 lb)(8 in.) – (1 lb)(12 in.) + (1.2 lb)(16 in.)  0 M A = 8.8 lb  in.

or M A = 8.8 lb in.

Page 5 of 7

BME 110A

Problem Set #3

5. Two steel pipes AB and BC each having a weight per unit length of 5 lb/ft are welded together at B and are supported by three wires attached at A, D, and C. Knowing that a  1.25 ft, determine the tension in each wire.

Page 6 of 7

Problem Set #3

BME 110A Free-Body Diagram:

W AB   5 lb/ft  2 ft   10 lb

First note

WBC   5 lb/ft  4 ft   20 lb

W  W AB  W BC  30 lb

To locate the equivalent force of the pipe assembly weight rG/B  W    ri  Wi   rG  AB   WAB  rG  BC   WBC

 xG i  zG k   30 lb j   1 ft k   10 lb j  2 ft i   20 lb j

or



 30 lb xG k  30 lb  zG i  10 lb ft  i  40 lb ft  k

From i-coefficient

zG 

10 lb  ft 1  ft 30 lb 3

k-coefficient

xG 

40 lb ft 1  1 ft 30 lb 3

From free-body diagram. of piping

M x  0:

W  zG   TA  2 ft   0 1  1   T A   ft  30 lb ft   5 lb 2  3 

Fy  0:

5 lb  TD  TC  30 lb  0  TD  TC  25 lb (1)

Page 7 of 7

or

TA  5.00 lb...