F1 = 3 kN F2 = 11 kN e-60 30 F3 2 kN 2 m 7 m F4 = 20 kN a 30° F5 = 2 kN Activate Windows 5] What is the moment of the force F5 about point B? Go to Settings to activate W d)-9.1 kN.m e) -11.1 kN.m 0-2.3 kN.m a) - 19.7 kN.m b)-6.7 kN.m c) - 4.8 kN.m 2 m 5 m
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- F1- 10 kN F2- 26 kN - 45 F3-8 KN 0-45 4 m 8m F4 = 44 kN F5 - 8 KN 1] What is the resultant moment of the five forces acting on the rod about point A? b) 108.1 kN.m d) 153.1 kN.m ) 23.1 KN.m ) 86.4 kN.m a) 77.5 kN.m ) 59.8 kN.m 2] What is the resultant moment of the five forces acting on the rod about point C? a) 44.5 kN.m b) 187.1 kN.m e) 59.8 kN.m d) 153.4 kN.m ) 99.1 kN.m O 86.4 kN.m 31 What is the moment of the force F1 about point E? a) - 14.6 kN.m b) - 106.1 kN.m ) - 48.8 kN.m d) - 215.8 kN.m ) - 99.7 kN.m 0-188.1 kN.m 41 What is the moment of the force F2 about point D? a) 19.5 kN.m b) 122.1 kN.m e) 88.8 kN.m d) 147.1 kN.m ) 58.1 kN.m ) 135.4 kN.m 5] What is the moment of the force F5 about point A? e) 44.8 kN.m a) 11.8 kN.m b) 5.7 kN.m d) 19.1 kN.m e) 69.4 kN.m ) 1.4 kN.m0.5m Q6) Newton's 2nd Law of Rotation and Angular Momentum 4.8m/s² 2kg A string is wrapped around a heavy wheel of radius R = 0.5 m. A block of mass m = 2 kg hangs from the other end of the string. When let go, the block falls with an acceleration of a = 4.8 m/s². The wheel rotates about its center with angular acceleration a, but otherwise does not move. The string does not slip. a) Draw a free body diagram and write Newton's 2nd law for the block Do not substitute any numbers. b) Draw a free body diagram and write Newton's 2nd law for the wheel. Do not substitute any numbers. c) What is the tension in the string, T? Write your answer to two decimal places d) What is the angular acceleration, a? Write your answer to two decimal places. e) What is the moment of inertia of the heavy wheel? Write your answer to two decimal places. f) The ice caps are found near the North and South poles of the Earth. If the ice caps were to melt, the water would spread out across all the oceans. What…F1 =5 kN F2 13 kN F3-4 kN B. 7m F4- 22 kN El - 30 FS = 4 kN 1] What is the resultant moment of the five forces acting on the rod about point B? a) 13.1 KN.m b) 54 kN.m ) 16.2 kN.m d) 18.5 KN.m e) 23.6 kN.m 09.1 KN.m 21 What is the resultant moment of the five forces acting on the rod about point E? a) - 11.7 KN.m b) -19.1 kN.m c) - 33 kN.m d) - 22.1 kN.m e) - 12.7 kN.m 9- 31.3 kN.m 3] What is the moment of the force F1 about point D? a) - 31.8 KN.m b) -19.1 kN.m c) - 3.3 KN.m d) - 22.1 kN.m e) - 12.7 kN.m 9- 9.3 kN.m 4] What is the moment of the force F2 about point E? a) 55,1 kN.m b) 69.3 kN.m e) 16.2 kN.m d) 18.5 kN.m e) 45.6 kNm f)75.1 kN.m Activate Windows 5] What is the moment of the force F5 about point B? c) -8.3 kN.m Go to Settings to activate We e) - 12.7 kN.m a) - 13.3 KN.m b) -17.1 kN.m d) - 22.1 kN.m )- 25.3 kN.m
- 2. Determine the moment of the four forces acting on the rod as in the fig. about point 'o' 'o' N OF N 07 N 09 N OS6. For each case illustrated in Figure below, determine the moment of the force about point 0. F-30 lb 60 F = 50 lb 4-0 45 F-10 Ib F= 40 lbWhiteboard Problem 2 object rotates, if it is A barbell consists of a 4 kg mass and a 12 kg mass separated by 0.5 m. 4 kg 105pxρώμ. ΑΣΠΡΟ 0.5 m 12 kg ripe caurer og ud 2 BLACu p (a) Relative to the 4 kg mass, where is the center of mass of the barbell? (b) Circular motion review: The barbell is rotating at 15 rpm about its center of mass At what linear speed (in m/s) is each mass moving?
- 0- 60° B - 30° F3 = 2 kN A 2 m 7 m F4 = 20 kN E a = 30° F5 = 2 kN 1] What is the resultant moment of the five forces acting on the rod about point B? a) 17.8 kN.m b) 20.4 kN.m c) 10.6 kN.m d) 12.9 kN.m e) 27.1 kN.m f) 30.1 kN.m 2] What is the resultant moment of the five forces acting on the rod about point E? a) – 21.7 kN.m b) -11.1 kN.m c) - 4.8 kN.m d) - 9.1 kN.m e) - 17.1 kN.m f) -27.3 kN.m 3] What is the moment of the force F1 about point D? a) - 5.8 kN.m b) - 21.2 kN.m c) - 25.6 kN.m d) - 9.9 kN.m e) 1- 8.1 kN.m f) - 26.1 kN.m 4] What is the moment of the force F2 about point E? a) 66.8 kN.m b) 25.4 kN.m c) 43.6 kN.m d) 62.9 kN.m e) 58.6 kN.m f) 23.1 kN.m Activate Windows 5] What is the moment of the force F5 about point B? Go to Settings to activate Wir e) - 11.1 kN.m а)-19.7 kN.m b) -6.7 kN.m c) - 4.8 kN.m d) - 9.1 kN.m f) -2.3 kN.mQ4) Determine the moment of force F about point O in fig (4). Express the result as a Cartesian vector. * F=300N 5m 2m FIG (4)Depicted below is a cantilever beam that is being subjected to a 13.5 N Force Couple. 13.54 13,5N 7a) What is Magnitude of the Moment about point A if L = 6.0 meters and d = 2.25 meters? Magnitude = N-m. 7b) Does the couple cause a positive Moment about point A, or does it cause a negative Moment about point A? Which one - Positive or Negative? 7c) What is Magnitude of the Moment about point A if L was increased to 8.8 meters instead of 6.0 meters but d remained equal to 2.25 meters? Magnitude = N-m.
- M L1 12 F2 L3 F3 L1=3.9m, L2=9.7m, L3=6.3m F1=8.8N, F2=5.8N, F3=7.4N calculate the sum of moments the weight of the beam is neglected and assume that the counterclockwise direction is positive.ARNING SYSTEM (ACADEMIC) eral Physics I- Spring21 Time left 0:16:30 Force F = -3.5 Ni+-7.2 Nj acts on a small rock with a position vector r=4 mi+6 mjrelative to the origin. In unit vector notation, what is the resulting torque on the rock about the point (9.7,4.5,7.4)m? O a. -49.8 j Nm b. -7.8 (-k) Nm O c. -57.2j Nm O d. -7.8 k Nm NEXT PAGETriceps muscle 4.0 cm Biceps muscle m₁ = 2.5 kg - 16 cm CG -X W 38 cm M (positive) "counter-clockwise" PHYSICS Determine the force generated by the biceps in order to maintain the forearm in this static position (zero movement). mo is the mass of the book. ma is the mass of the forearm/wrist/hand segment. The appropriate moment arms are indicated. +y m, = 4.0 kg CG Wo M (negative) "clockwise" +X