statics 4

statics 4 - Problem 4.1 Determine the moment of the 50-N...

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Problem 4.1 Determine the moment of the 50-N force about (a) point A , (b) point B . 3 m 5 m B P A 50 N Solution: Use 2-dimensional moment strategy: determine nor- mal distance to line of action D ; calculate magnitude DF ; deter- mine sign. (a) The perpendicular distance from A to line of action is D =0 , hence the moment M A = DF . (b) The perpendicular distance from B to line of action is 3 m (the triangle is a 3-4-5 right triangle), and the action is counter clock- wise, hence M B = +(3)(50) = +150 N-m. A B P F 5 m 50 N 3 m Problem 4.2 The radius of the pulley is r . 2 m and it is not free to rotate. The magnitudes of the forces are | F A | = 140 N and | F B | = 180 N. (a) What is the moment about the center of the pulley due to the force F A ? (b) What is the sum of the moments about the center of the pulley due to the forces F A and F B ? 45 F A F B 10 ° r ° Solution: ± + M A = r | F A | =(0 . 2)140 N-m M A =28 N-m ± + M B = r | F B | = (0 . 2)180 ± + M B = 36 N-m ± + M A + M B 36 = 8 Nm ± +( M A + M B )= 8 N-m 45 ° F A F B 10 ° r r r F B F A 45 ° 10 °
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Problem 4.3 The wheels of the overhead crane exert downward forces on the horizontal I-beam at B and C . If the force at B is 40 kip and the force at C is 44 kip, determine the sum of the moments of the forces on the beam about (a) point A , (b) point D . A BC D 25 ft 10 ft 15 ft Solution: Use 2-dimensional moment strategy: determine nor- mal distance to line of action D ; calculate magnitude DF ; determine sign. Add moments. (a) The normal distances from A to the lines of action are D AB = 10 ft, and D AC =35 ft. The moments are clockwise (negative). Hence, X M A = 10(40) 35(44) = 1940 ft-kip . (b) The normal distances from D to the lines of action are D DB = 40 ft, and D DC =15 ft. The actions are positive; hence X M D = +(40)(40) + (15)(44) = 2260 ft-kip A D 10 ft 25 ft 15 ft Problem 4.4 If you exert a 90-N force on the wrench in the direction shown, what moment do you exert about the center of the nut? Compare your answer to the moment exerted if you exert the 90-N force perpendicular to the shaft of the wrench. 90 N 500 mm 450 mm Solution: M = d 1 · F =( . 45) 90=40 . 5 N-m clockwise for direction shown M P = d 2 · F =(0 . 5) 90=45 N-m clockwise for perpendicular force 500 mm 90 N
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Problem 4.5 If you exert a force F on the wrench in the direction shown and a 50 N-m moment is required to loosen the nut, what force F must you apply? F 300 mm 265 mm Nut F Solution: Use 2-dimensional moment strategy: determine nor- mal distance to line of action D ; calculate magnitude DF ; determine sign. Solve for unknown force. The normal distance from the nut center to the line of action is D = 0 . 265 m. Thus to loosen the nut, 50 = D | F | =0 . 265 | F | . Solve: | F | = 50 0 . 265 = 188 . 7 N in the direction shown. 265 mm 300 mm F Nut Problem 4.6 The support at the left end of the beam will fail if the moment about P due to the 20-kN force exceeds 35 kN-m. Based on this criterion, what is the maximum safe value of the angle α in the range 0 α 90 ?
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This note was uploaded on 11/02/2008 for the course MAE 101 taught by Professor Orient during the Spring '08 term at UCLA.

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statics 4 - Problem 4.1 Determine the moment of the 50-N...

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