Moment

226. In Fig. P-226 assuming clockwise moments as positive, compute the moment of force F = 450 lb and of force P = 361 lb about points A, B, C, and D.

A F

C

1’ D

P B

1’

FIG. P-226.

MA = Fy (x) – Fx (y)

MC = Fy (x) – Fx (y)

MA = -Fy (1) – Fx (3)

MC = Fy (5)

3 4 MA = 450   1  450    3 5 5

3 MC = 450    5  5

M A   1350 lb  ft

M C  1350 lb  ft

MB = Fy (x) - Fx (y)

MD = Fy (x) – Fx (y)

MB = Fy (4) + Fx (3)

MD = -Fy (1) + Fx (3)

3 4 MB = 450    4   450    3 5 5

3 4 MD = 450   1  450    3 5 5

M B  2160 lb  ft

M D  810 lb  ft

MA = Py (x) – Px (y)

MB = -Py (3) – Px (3)

MA = Py (2) – Px (3)

 3   2  MB = 361   3  361   3  13   13 

 3   2  MA = 361   2   361   3  13   13 

M B  300.37 lb  ft

MA = 0

MB = Py (x) – Px (y)

227. Two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. Force P is 200 lb directed up to the right at 30° with the horizontal and force Q is 100 lb directed up to the left at 60° with the horizontal. Determine the moment of the resultant of these two forces with respect to O. Y

Y

60°

30°

3 ft 4 ft

X

F

 200lb  cos30   100lb  cos 60

F

 P cos  Q cos

F

 P sin   Q sin 

F

 123.21lb

F

 200lb  sin 30   100lb  sin 60

F

 186.60lb

x

x

x

 y

y

y



MO = Fx  d   Fy  d  MO = 123.21 (3) – 186.60 (4) MO = - 376.77 counterclockwise

228. Without computing the magnitude of the resultant, determine where the resultant of the forces shown in Fig. P-228 intersects the X and Y axes.

Y

5’’

500 lb

5’’

X

O

361 lb FIG. P-228.

F

x

 F cos 

 3   3   500lb    361lb    18   18   Fx  653.92lb

F

x

M M M

O

 Fx d   Fy d 

O

 653.92lb  2 ft   153.31lb  2 ft 

O

 1001.23 ft  lb

F

y

 F sin 

 3   2   500lb    361lb    18   18   Fy  153.31lb

F

y

M

O

M

 Fy ix

O

 Fxiy

1001.23 ft  lb  153.31lb  ix 

1001.23 ft  lb  653.92lb  ix 

1001.23 ft  lb  ix 153.31lb ix  6.53 ft 

1001.23 ft  lb  ix 653.92lb ix  1.53 

229. In Fig. P-229, find the y coordinate of point A so that the 361-lb force will have clockwise moment of 400 ft-lb about O. Also determine the X and Y intercepts of the action line of the force.

Y

361 lb 2 3

XA = 2 ft A

YA

X O FIG. P-229.

Fx  cos  Fy  sin 

 3  Fx  361   13  Fx  300.37lb iy 

400 ft  lb 300.37lb

 2  Fy  361   13  Fy  200.25lb

M O  FxYA  Fy X A

400 ft  lb  300.37lb YA   200.25lb  2 ft  300.37lb YA   400 ft  lb  400.50 ft  lb YA 

800.50 ft  lb 300.37lb

YA  2.67 ft

M O  Fx i y

iy 

iy 

MO Fx

400 ft  lb 300.37lb

ix 

MO Fy

ix 

400 ft  lb 200.25lb

ix  2 ft (left of O)

i y  1.33 ft (above O)

M O  Fy ix

230. For the truss shown in Fig. P-230, compute the perpendicular distance from E and from G to the line BD. Hint: Imagine force F directed along BD and compute its moment in terms of its components about E and about G. Then equate these results to the definition of moment M = Fd to compute the required perpendicular distances.

F D B 16’

16’

18’

12’

12’

A 12’

C

12’

E

4 12

G

12’

Fx Fy F   12 4 160 Fx F  12 160 Fx 

About E

M

E

M

E

M

E

 Fd  F 16 

12 F 16   Fd  160

15.18 ft  d 

About G

12’

F

D

B

12’

12 F 160

M

G

 Fd

M

G

 F  20

M

G



12 F  20   Fd  160

18.97 ft  d 

12’

Y

A 3’ X O

6’

B

FIG. P-231, P-232, and P-233.

231. A force P passing through points A and B in Fig. P-231 has a clockwise moment of 300 ft-lb about O. Compute the value of P. Y

A 3’ X O

M O  Px d 

300 ft  lb  Px  3 ft  Px 

300 ft  lb 3 ft

6’

B

P

300 ft  lb  Py  6 ft  Py 

300 ft  lb 6 ft

Py  50lb

Px  100lb M O  Py d 

P  Py 2  Px 2

P

 50lb   100lb  2

2

P  111.8lb tan  

 50lb    100lb 

  tan 1 

Py

  26.57

Px

232. In Fig. 231, the moment of a certain force F is 180 ft-lb clockwise about O and 90 ft-lb counterclockwise about B. If its moment about A is zero, determine the force.

Y

A 3’ X O

6’

B

F

90  60 3  Fy  6

M O  Fx d 

180  Fx  3

270   Fy 6

180  Fx 3

45lb  Fy

60lb  Fx

M A  Fx  y   Fy  x   45lb    60lb 

F  Fy 2  Fx 2

F

 45   60  2

F  75lb tan  

Fy Fx

  tan 1  2

  36.87

233. In Fig. P-231, a force P intersects the X axis at 4 ft to the right of O. If its moment about A is 170 ft-lb counterclockwise and its moment about B is 40 ft-lb clockwise, determine its y intercept. Y

A 3’

4’ O

X 6’

B

M B  Py d 

M A  Px  3  Py  4

40 ft  lb  Py  2 ft 

170 ft  lb  Px 3 ft   20lb  4 ft 

40 ft  lb 2 ft

170 ft  lb  80 ft  lb  Px 3 ft 

Py 

Py  20lb

90 ft  lb  Px  3 ft  Px 

90 ft  lb 3 ft

Px  30lb

M A  Px  y   Py  x 

M O  Py d 

M O  Px i y

MO   20lb 4 ft 

80 ft  lb  30lb  iy 

M O  80 ft  lb

80 ft  lb  iy 30lb 2.67 ft  i y