Hw 5 - A 200-N force is applied as shown to the bracket...

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PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, detennine the moment of the 16-N force about Point B by resolving the force into horizontal and vertical components. SOLUTION Note that f} =a-2Do = 28"_20° =8" and F~ = (16 N)cos8 D = 15.&443 N F x = (16 N)sin 8" = 2.2268 N Also x = (0.17 m)cos20° = 0.159748 m y = (0.17 m)sin20° = 0.058143 m Noting that the direction of the moment of eaeh force component about B is counterclockwise. M B =xF y + yF x =(0.159748 rnX2.2268 N) +(0.058143 rn)(I5.8443 N) =1.277N·m or M B =l.277N.m) . ....
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PROBLEM 3.2 A foot valve for II pneumatie system is hinged at B. Knowing that a = 28<l, detennine the moment of the 16-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC. SOLUTION First resolve the 4-lb force into components P and Q. where Q=(16 N)sin2BO =7.5115N Then M B = rA1BQ = (0.17 m)(7.5115 N) =1.277N-m
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PROBLEM 3.21
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Unformatted text preview: A 200-N force is applied as shown to the bracket ABC. Detennine the moment ofthe force about A. SOLUTION We have where [&quot;CIA = (0.06 m)1 + (0.075 m)j Fe = -(200 N)cos 30&quot;j + (200 N)sin 30&quot;k Then I M A ~ 200 0.06 o j 0.Q75 -cos30&quot; k 0 sin 30&quot; ~ 200[(0.075,io 30)1 -(0.06';0 30 0 )j - (0.06co, 30 0 )k] or M. ~(7.50Nm)i-(6.00Nm)j-(i039Nm)k &lt;II PROBLEM 3.26 A small boat hangs from two davits, one of whieh is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant foree R A exerted on the davit at A. SOLUTION We have where and Thus Also U,ing Eq. (3.21), R A =2F,(8+F AD F,a =-(821b)j F =F AD=(82Ib)61-7.75 1-3k AD AD AD 10.25 F'D = (48 Ib)i - (62 Ib)J -(24Ib)k R, = 2F,. + F'D = (48 Ib)l- (226 Ib)J- (24Ib)k r&quot;e =(7.75 ft)j+(3 ft)k i j k Me = 0 7.75 3 48 -226 -24 = (492 lb ft)1 + (1441b ft)j - (372 lb ft)k Me =(492 lb ft)1 + (144.0 Ibft)J-(372lbft)k . .....
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Hw 5 - A 200-N force is applied as shown to the bracket...

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