5 - Problem 5.1 The beam has pin and roller supports and is...

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Problem 5.1 The beam has pin and roller supports and is subjected to a 4-kN load. (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports. Strategy: (a) Draw a diagram of the beam iso- lated from its supports. Complete the free-body di- agram of the beam by adding the 4-kN load and the reactions due to the pin and roller supports (see Ta- ble 5.1). (b) Use the scalar equilibrium equations (5.4)–(5.6) to determine the reactions. 4 kN B A 2 m 3 m Solution: X F X =0 : A X X F Y : A Y + B Y 4 kN X M A : 2(4 kN )+3 B Y B Y =8 / 3 kN A Y =4 / 3 kN A X ,A Y =1 . 33 kN ,B Y =2 . 67 kN 4 kN B A 2 m 3 m 3 m 2 m A X B X A Y B Y 4 kN Problem 5.2 The beam has a built-in support and is loaded by a 2-kN force and a 6 kN-m couple. (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports. 2 kN 6 kN-m 2 m 3 m A Solution: (a) ( b ) X F X : A X X F Y : A Y 2 kN X M A : M A (2)(2 kN )+6 kN-m M A = 2 kNm A X A Y kN 2 kN 6 kN-m 2 m 3 m A A X y x A Y 3 m 2 m 2 kN 6 kN-m M A
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Problem 5.3 The beam is subjected to a load F = 400 N and is supported by the rope and the smooth sur- faces at A and B . (a) Draw the free-body diagram of the beam. (b) What are the magnitudes of the reactions at A and B ? F 30 ° 45 ° AB 1.2 m 1.5 m 1 m Solution: X F X =0 : A cos 45 B sin 30 X F Y : A sin 45 + B cos 30 T 400 N ± + X M A : 1 . 2 T 2 . 7(400) + 3 . 7 B cos 30 Solving, we get A = 271 N B = 383 N T = 124 N F 30 ° 45 ° 1.2 m 1.5 m 1 m x y 1.2 m 1.5 m 45 ° 30 ° T 1 m B F A Problem 5.4 (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports. 3 m B A 5 kN 3 m Solution: (a) ( b ) X F X : A X X F Y : A Y + B Y 5 kN ± + X M A : 3 B Y 6(5 kN )=0 Solving: A X B Y =10 kN A Y = 5 kN 3 m B A 5 kN 3 m A X A Y B Y 5 kN x y 3 m 3 m
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Problem 5.5 (a) Draw the free-body diagram of the 60-lb drill press, assuming that the surfaces at A and B are smooth. (b) Determine the reactions at A and B . 60 lb AB 10 in 14 in Solution: The system is in equilibrium. (a) The free body diagram is shown. (b) The sum of the forces: X F X =0 , X F Y = F A + F B 60=0 The sum of the moments about point A : X M A = 10(60) + 24( F B )=0 , from which F B = 600 24 =25 lb Substitute into the force balance equation: F A =60 F B =35 lb 60 lb 10 in 14 in F A F B
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Problem 5.6 The masses of the person and the diving board are 54 kg and 36 kg, respectively. Assume that they are in equilibrium. (a) Draw the free-body diagram of the diving board. (b) Determine the reactions at the supports A and B . W P AB W D 1.2 m 2.4 m 4.6 m Solution: (a) ( b )( N ) X F X =0 : A X ( N ) X F Y : A Y + B Y (54)(9 . 81) 36(9 . 81)=0 X M A : 1 . 2 B Y (2 . 4)(36)(9 . 81) (4 . 6)(54)(9 . Solving: A X N A Y = 1 . 85 kN B Y =2 . 74 kN W P W D 1.2 m 2.4 m 4.6 m A Y A X B Y W P W D 1.2 m 2.4 m 4.6 m Problem 5.7 The ironing board has supports at A and B that can be modeled as roller supports. (a) Draw the free-body diagram of the ironing board.
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This note was uploaded on 04/03/2008 for the course MAE 101 taught by Professor Orient during the Winter '08 term at UCLA.

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5 - Problem 5.1 The beam has pin and roller supports and is...

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