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HW1 - PROBLEM 2.3'l‘wo forces P and Q are applied as...

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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 u-nr' and b—h'. [(1) Using Irlgononmry. dclcrmine the angle a knowingI that the componem "long b-b’ is 30 lb. (/7) What is the corresponding value of '- ' 111a component along u-u"? 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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