RSL-2 A rod of negligible mass is pulled downat each end as shown to the right. The rodis supported by a rope as shown.a) Draw the free body diagram for the rodon a figure like the one to the right.support ropeb) Find the net torque on the rod about thepoint where the rope supports it.2ft5ftc) Which way does the rod tend to rotate,clockwise or counterclockwise?35lb15lbd) Which way does the angular acceleration vectora point?99+bp

Answered: Image /qna-images/answer/55103c57-b804-411d-8a14-11dbc05e4659.jpg

Answered: RSL-2 A rod of negligible mass is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point P
The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. d 2 Pl CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m С. 62,5 N*m D. Object 1 doesn't exert a torque.
Problem C) A square plate (4 meters on each side) is fixed from a pivot point and can rotate around this point as shown below. Each of the three forces are applied perpendicular to their respective side. Find the total torque applied by the given forces and state which direction the plate will rotate (clockwise or counter-clockwise). Pivot Point 3m 4 m F, = 3N 1m 2 m 1m F2 = 1N, F3 = 5N 2 m 3m
Four forces, FA, FB, FC, and FD, each of magnitude 1 N, are applied to a square object at the points shown below (at the tails of the arrows), and each produces a torque around the pivot (the circle). Rank these torques, τA, τB, τC , and τD, from most negative to most positive. Use the sign convention for torques. a)τA < τD < τC < τB b)τA < τD < τB < τC c)τC < τD < τB < τA d)τB < τC < τA < τD e)τD < τB < τA < τC
You must solve all questions. The aluminum bar has a 40mm diameter and 600mm length. It is held stationary at the origin on the left end. On the right end is a wrench that is 70i+220j mm from the connection to the rod to where the force is applied. At that point, a force of -7000i+800j N is applied. a) What is the mass of the bar? b) What is the moment applied by the wrench? c) What is the maximum normal stress due to this moment? d) What is the yield factor of safety for this stress?
(٢٦+ * Find the two forces Fbc & Fac as :shown in the figure below Fab=30 N 110 30/Fbc (Fbc=45.18 N and Fac= 58.8 N) O (Fbc=38.56 N and Fac= 56.38N) O (Fbc=75.18 N and Fac= 78.8 N) O ! Fac
Draw the Free Body Diagram Solve step by step Always used 3 decimal places! Thank you in advance
The shafts of an angle drive are subjected to the action of the two torques shown in the figure. Replace both torques by a single a) equivalent torque and b) specify its magnitude and direction. S lb-ft. 6 lb-ft
F1 F2 A F4 Consider the following values: - F5 a -7 m; b - 3 m; a -7 m; d - 3 m; F1 3 kN; F2 - 11 KN; F3 - 2 kN; F4 - 20 kN; F5 - 2 kN; a= 30°, e= 60°, B= 30° 1) What is the resultant moment of the five forces acting on the rod about point A? a) 90.8 KN.m m.220ט e) 14.6 KN.m d) 55.9 kN.m e) 21.6 kN.m H 61.3 KN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 77.8 KN.m b) 25.7 kN.m ) 38.6 kN.m d) 18.9 KN.m e) 37.9 KN.m ) 84.2 KN.m 3] What is the resultant moment of the five forces acting on the rod about point C? )35.7 kN.m b) 21.1 KN.m ) 714.6 kN.m d) 69.1 kN.m *) SL4 kN.m 0 85.3 N.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 17.8 KN.m b) 55.7 kN.m ) 44.6 kN.m d) 18.9 kN.m e) 21.5 kN.m 0 30.1 kN.m 5] What is the moment of the force F2 about point E? ) 64.6 kN.m a) 81.7 kN.m b) 78.1 kN.am d) 73.1 KN.m e) 54.6 kN.m H 61.3 kN.m
5. A pull in a bicycle chain of 250 N applied to a sprocket of diameter 80 mm provides the necessary torque to the rear wheel. If the wheel is 600 mm in diameter, what is the force between the wheel and the ground?
F1 F2 a F3 F4 Consider the following values: - F5 a = 7 m; b = 3 m; c 7 m; d = 3 m; F1 = 3 kN; P2 11 kN; F3 = 2 kN; F4 = 20 kN; P5 2 kN; a = 30°, e= 60°, B= 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) 90.8 KN.m ) 146 KN.m d) 55.9 KN.m H 61.3 KN.m b) 72.7 KN.m e) 21.6 KN.m
For a Davis steering gear, the gear's track arm is at a distance of 160 mm from the front main axle and the difference between their lengths is 100 mm. If the distance between steering pivots of the main axle is 2 m a) find the inclination of the track arms to the longitudinal axis of the vehicle b)determine the length of the chassis between the front and the rear wheels.

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