3.152 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. -x 24 in. D 17 in. 30 in. Q H 32 in. 16 in. 18 in. 21 in. Fig. P3.152 B x 12 in. Substitute (32 î – 30 j – 24k) in for AB in the above equation to calculate A. = λ = (321-30ĵ-24k) in 32²+(-30)²+(-24)² 321-30 j-24 k 50 0.64 î – 0.6 ĵ – 0.48 k (1) Write the expression for the vector from B to H. TBH = (−32 î+17 ĵ) in . (II) Write the expression for the vector from D to H. DH = (-16-21-12k) in == Write the expression to calculate the force Q in the direction of the vector DH. Q = |Q| DH DH Here is the magnitude of force Q. Substitute 174 lb for |Q| and (−16 î – 21 ĵĴ – 12 k) in for DH in the above equation to calculate Q. Q = (174 lb) = 174 lb 29 in (-16 1-21 j-12k) in √√(−16)²+(-21)²+(-12)² in -16 î – 21 ĵ – 12k) in = (−96 î – 126 } – 72k) lb .(111) -
3.152 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. -x 24 in. D 17 in. 30 in. Q H 32 in. 16 in. 18 in. 21 in. Fig. P3.152 B x 12 in. Substitute (32 î – 30 j – 24k) in for AB in the above equation to calculate A. = λ = (321-30ĵ-24k) in 32²+(-30)²+(-24)² 321-30 j-24 k 50 0.64 î – 0.6 ĵ – 0.48 k (1) Write the expression for the vector from B to H. TBH = (−32 î+17 ĵ) in . (II) Write the expression for the vector from D to H. DH = (-16-21-12k) in == Write the expression to calculate the force Q in the direction of the vector DH. Q = |Q| DH DH Here is the magnitude of force Q. Substitute 174 lb for |Q| and (−16 î – 21 ĵĴ – 12 k) in for DH in the above equation to calculate Q. Q = (174 lb) = 174 lb 29 in (-16 1-21 j-12k) in √√(−16)²+(-21)²+(-12)² in -16 î – 21 ĵ – 12k) in = (−96 î – 126 } – 72k) lb .(111) -
3.152 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. -x 24 in. D 17 in. 30 in. Q H 32 in. 16 in. 18 in. 21 in. Fig. P3.152 B x 12 in. Substitute (32 î – 30 j – 24k) in for AB in the above equation to calculate A. = λ = (321-30ĵ-24k) in 32²+(-30)²+(-24)² 321-30 j-24 k 50 0.64 î – 0.6 ĵ – 0.48 k (1) Write the expression for the vector from B to H. TBH = (−32 î+17 ĵ) in . (II) Write the expression for the vector from D to H. DH = (-16-21-12k) in == Write the expression to calculate the force Q in the direction of the vector DH. Q = |Q| DH DH Here is the magnitude of force Q. Substitute 174 lb for |Q| and (−16 î – 21 ĵĴ – 12 k) in for DH in the above equation to calculate Q. Q = (174 lb) = 174 lb 29 in (-16 1-21 j-12k) in √√(−16)²+(-21)²+(-12)² in -16 î – 21 ĵ – 12k) in = (−96 î – 126 } – 72k) lb .(111) -
Transcribed Image Text:3.152 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in.
rod AB. Determine the moment about AB of the 174-lb force Q.
-x
24 in.
D
17 in.
30 in.
Q H
32 in.
16 in.
18 in.
21 in.
Fig. P3.152
B
x
12 in.
Transcribed Image Text:Substitute (32 î – 30 j – 24k) in for AB in the above equation to calculate A.
=
λ =
(321-30ĵ-24k) in
32²+(-30)²+(-24)²
321-30 j-24 k
50
0.64 î – 0.6 ĵ – 0.48 k
(1)
Write the expression for the vector from B to H.
TBH = (−32 î+17 ĵ) in . (II)
Write the expression for the vector from D to H.
DH = (-16-21-12k) in
==
Write the expression to calculate the force Q in the direction of the vector DH.
Q = |Q|
DH
DH
Here is the magnitude of force Q.
Substitute 174 lb for |Q| and (−16 î – 21 ĵĴ – 12 k) in for DH in the above equation to calculate Q.
Q = (174 lb)
=
174 lb
29 in
(-16 1-21 j-12k) in
√√(−16)²+(-21)²+(-12)² in
-16 î – 21 ĵ – 12k) in
= (−96 î – 126 } – 72k) lb
.(111)
-
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
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