Week 4 lecture - ubc-phys170 PDF

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PROBLEM 4-53 (page 146, 13th edition) PROBLEM 4-55 (page 145, 12th edition)

! • Determine the moment of F about A. ! • Determine the magnitude of the moment of F about an axis extending from A to C.

4-53 (page 146, 13th edition)

4-55 (page 145, 12th edition)

COMPLETION OF PROBLEM MOMENT OF THE FORCE

! ! • Moment M A of F about A (suppressing units): ! ! ! i j k ! ! ! M A = rAB ! F = 4 3 !2 4 12 !3 ! ! ! = i [3(!3) ! 12(!2)] ! j [4(!3) ! 4(!2)] + k [4(12) ! 4(3)]

! ! ! ! M A = (15 i + 4 j + 36 k) lb!ft ! Magnitude and coordinate direction angles of M A : M A = 39.2 lb!ft ! = 67.5o

! • Magnitude of F : ! Lever arm for F :

! = 84.1o

! = 23.2o

F = 13.0 lb d = M A / F = 3.02 ft

4-53 (page 146, 13th edition)

4-55 (page 145, 12th edition)

COMPLETION OF PROBLEM MOMENT OF THE FORCE ABOUT AN AXIS

! • Magnitude M AC of the moment of F about AC : M AC

! ! = uAC i M A

! ! ! ! ! ! 4i + 3 j r uAC = AC = = 0.8i + 0.6 j rAC 4 2 + 32

! ! ! ! M A = (15 i + 4 j + 36 k) lb!ft M AC = 14.4 lb!ft

! • Alternative method (without first calculating M A ): M AC

! ! ! ! ! = uAC i M A = uAC i (rAB ! F)

! ! ! uAC = 0.8i + 0.6 j ! ! ! ! rAB = (4 i + 3 j ! 2 k)ft ! ! ! ! F = ( 4 i + 12 j ! 3k) lb

M AC

0.8 0.6 0 = 4 3 !2 lb"ft = 14.4 lb"ft 4 12 !3

PROBLEM 4-116 (page 169 12e) (See Problem 4.116. jpg) ! ! The pipe assembly is acted on by forces F1 and F2 and by a couple ! ! ! ! moment M = (5 i + 6 j + 7 k) lb!ft . ! • Determine the magnitude and coordinate direction angles of M .

• Determine the magnitude of each of the two forces comprising the couple when the lever arm of the couple is 0.5 ft.

• Replace the force-couple system by a resultant force and couple moment at O. Express the results in Cartesian vector form.

4-116 (page 169, 12th edition) COMPLETION OF PROBLEM

! • Magnitude and coordinate direction angles of M : M = 5 2 +6 2 +7 2 lb ft = 10.5 lb ft ! = cos"1(5 /10.49) = 61.5!

! = cos"1(6 /10.49) = 55.1! ! = cos"1(7 /10.49) = 48.1!

• When d = 0.5 ft :

F = M / d = 10.5 / 0.5 lb = 21.0 lb ! • Resultant force FR at O: ! ! ! ! ! FR = ! F = ("30.0 i + 15.0 j + 45.0 k) lb ! Resultant couple moment (M R )O at O: ! ! ! ! (M R )O = ! M + ! ( r " F) ! i

! j

! k

! i

! j

! k

! ! ! = ( 5 i + 6 j + 7 k + 1.5 2 0 + 1.5 4 2 ) lb ft !20 !10 25 !10 25 20 ! ! ! = (85.0 i ! 81.5 j + 110 k) lb"ft

PROBLEM 4-137 (page 182, 13th edition) PROBLEM 4-141 (page 182, 12th edition) • Replace the three forces acting on the plate by a wrench. • Determine the magnitude of the force and couple moment of the wrench, and the point P(x,y,0) where its line of action intersects the plate.

4-137 (page 182, 13th edition)

4-141 (page 182, 12th edition)

COMPLETION OF PROBLEM (page 1) RESULTANT FORCE AND COUPLE MOMENT

! • Resultant force FR at point P (suppressing units): ! ! ! ! ! FR = ! F = 500 i + 300 j + 800 k ! • Resultant couple moment (M R )P at point P (suppressing units): ! ! ! ! (M R )P = ! M + ! (r " F)

=

! i !x

! ! ! ! i j j k ! y 0 + !x 4!y

500

0

0

0

0

! k 0 800

! ! ! i j k + 6!x 4!y 0 0

300

0

! ! ! = (3200 ! 800 y) i + 800x j + (500y +1800 ! 300x) k

4-137 (page 182, 13th edition)

4-141 (page 182, 12th edition)

COMPLETION OF PROBLEM (page 2) : WRENCH EQUATIONS ! ! ( M ) at point P: • Resultant force FR and couple moment R P

! ! ! ! FR = 500 i + 300 j + 800 k ! ! ! ! (M R )P = (3200 ! 800y) i + 800x j + (500y +1800 ! 300x) k

(1)

! ! ( M ) F • For a wrench, must be parallel (or antiparallel) to R . We write R P ! ! (M R )P = FR d where d is to be determined. ( d is the lever arm for the resultant couple when each force in the couple has magnitude FR .) That is,

! ! ! ! (M R )P = (500 i + 300 j + 800 k) d

• We determine equations for x, y, d by equating (1) and (2):

3200 ! 800 y = 500d 800 x = 300d 500 y +1800 ! 300 x = 800d Solving:

(x, y, d) = (1.16, 2.06, 3.10)

Re-inserting units:

FR = 990 N

(M R )P = 3.07 kN m

! = 59.7!

" = 72.4!

(x, y) = (1.16, 2.06)m # = 36.1!

(2)...