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Unformatted text preview: Problem 2.12 The rope ABC exerts forces FM and Solution: Draw the vectors accurately and then measure the Far: of equal magnitude on the block at B. The . unknown magnitudes.
magmtude of the total force exerted on the block by
the two forces is 200 lb. Graphically determine |F3A|. Ile = 174 lb Problem 2.49 The ﬁgure shows three forces acting on
a joint of a structure. The magnitude of FL. is 60 kN, and
FA + F 3 + F5 : 0. What are the magnitudes of FA and
F3? Solution: We need to write each force in terms of its components.
FA 2 IF,“ oos40i+ \FAlsin 40j (kN)
F}: = |F3\ COS 195°i + 3?}! sin l95j (kN) if = |Fcicos 270°i + \Fd sin270“j (kN)
Thus Fc : —60j kN Since FA + F3 + E: = 0, their components in each direction must also
am to zero. FM+FHX+FCX :0
FAy+FBy+FCy :0 lFAlcos40“ + |F5i cos 195° +0 = 0
‘ .IFAlsin40° + |F3|sin 195° — 60 (kN) : 0 ' g for iFA| and ing, we get "|=137 kN, \Fgl : 109 kN _'a--nlem 2.53 The three forces acting on the car are Solution:
u' . The force T is parallel to the x axis and the
itude of the force W is 14 kN. IfT + W + N = 0, "m are the magnitudes of the forces T and N? 2F; : T ‘NSin 20° = 0 2F).CNCOSZOD*14RN=U Solving we ﬁnd N 214.90 N, T = 5.l0 N ...
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- Spring '08