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# 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: F X = 0: A X = 0 F Y = 0: A Y + B Y 4 kN = 0 M A = 0: 2(4 kN ) + 3 B Y = 0 B Y = 8 / 3 kN A Y = 4 / 3 kN A X = 0 , 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 ) F X = 0: A X = 0 F Y = 0: A Y 2 kN = 0 M A = 0: M A (2)(2 kN ) + 6 kN-m = 0 M A = 2 kNm A X = 0 A Y = 2 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 ° A B 1.2 m 1.5 m 1 m Solution: F X = 0: A cos 45 B sin 30 = 0 F Y = 0: A sin 45 + B cos 30 T 400 N = 0 + M A = 0: 1 . 2 T 2 . 7(400) + 3 . 7 B cos 30 = 0 Solving, we get A = 271 N B = 383 N T = 124 N F 30 ° 45 ° A B 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 ) F X = 0: A X = 0 F Y = 0: A Y + B Y 5 kN = 0 + M A = 0: 3 B Y 6(5 kN ) = 0 Solving: A X = 0 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
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 A B 10 in 14 in Solution: The system is in equilibrium. (a) The free body diagram is shown. (b) The sum of the forces: F X = 0 , F Y = F A + F B 60 = 0 The sum of the moments about point A : 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 A B 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 A B W D 1.2 m 2.4 m 4.6 m Solution: (a) ( b ) ( N ) F X = 0: A X = 0 ( N ) F Y = 0: A Y + B Y (54)(9 . 81) 36(9 . 81) = 0 M A = 0: 1 . 2 B Y (2 . 4)(36)(9 . 81) (4 . 6)(54)(9 . 81) = 0 Solving: A X = 0 N A Y = 1 . 85 kN B Y = 2 . 74 kN W P A B 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.
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5 - Problem 5.1 The beam has pin and roller supports and is...

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