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Unformatted text preview: PROBLEM 2.3 'l‘wo forces P and Q are applied as shown at point .4 Mil hook suppori.
Knowing that P : 15 lb and Q ' 25 lb, determine graphically the
magnitude and direction oftheir resultant using (a) the parallelogram law.
(b) the triangle rule. SOLUTION (a) (b) We measure: R = 37 lb. (2 7 76'” PROBLEM 2.8 The SO—Ib Force is to be resolvud into componunls along linus unr' and
b—h'. [(1) Using Irlgononmry. dclcrmine the angle a knowingI that the
componem "long bb’ is 30 lb. (/7) What is the corresponding value of
' ' 111a component along uu"? 12‘ 11' SOLUTION Fvawh Using tho triangle rule and the Lav» ofslnes sin a sin 40” (a) 30% = 5011:
5111a .. (1.3857
c: 22.1“ 4
(in a + [1+ 40? = 1ch—
;3' =117V31°
5011: my sin ,1} Sin 4U” w
I 1
'ﬂ 5* F = 69,1“: 4 ”l1 I PROBLEM 2.12 For the hook support of Problem 2.3. using lrigunometry and knuwing
I that the magnitude of P is 35 lb, dclermine (a) the required magnitude of
the force 0 ii' the resultanl R of lhe tun forcex applied Lil .! is to be
verticai‘ (h) the corresponding magnitude ol'R. Problem 2.3: Two forces P and Q are applied as shown {II point x] 0ft}
hook support. Knouing that 1" : i5 lb and Q = 25 ih. determine
graphically the magnitude and direction ol‘theii' resultant using (a) the
paralleiugrzim Jim. [32) the li'iungle rule. SOLUTION Uxing the triangle mic“ and the Law oj'Sines (a) Haw: Q : 12,94 Ib‘
Ht) [3:18(J°—{15"‘¢30°)
= I35c
3"
Thus: :3 — : l)”: R = 25 ibi 5’1—‘ii = 35.361?)
i siiﬂﬂ“ t tT.‘ “” I ...
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 Spring '08
 Jenkins
 Statics

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