G. 620mm 1.8m 1.2m
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A: To determine the reaction at A and B, the beam can be dealt in two sections that are AD and DB. In…
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A: In step 1 we draw given diagram and calculate reaction forces:-
Q: Vertically 300lb/pulg2 and horizontally 150 lb/pulg2 I have only this data please answer.
A: Giver:- My=300 lb/pulg2 Mx=150 lb/pulg2 To find:- Resultant
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Q: -150 mm 15 mm B C 150 mm -15 mm 15 mm - 100 mm·
A: given;
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A: Method of Section: In this method, the reaction in the truss member is calculated by cutting the…
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A:
Q: 3 kN 12 kN.m 3 kN / m 5 kN / m 7 3 m 3 m 2 m
A: given:
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A: Dear student, we are suppose to solve only one question. Please post other question as a separate…
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A: Given Data : 250cm+10m
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A: A beam is defined as a structural member subjected to transverse shear load during its…
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A: Let RFx and RFy are the horizontal and vertical reaction at F RE is vertical reaction at E By using…
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A: Let us draw the reaction force
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A: The given is as shown below,
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A: Writing the co-ordinates of the points as follows, Writing the position vector of the line AB, AC…
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A: Givenρb=8.75 Mg/m3=8750 kg/m3ρs=7.87 Mg/m3=7870 kg/m3
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A: The reactions on the supports of beam can be drawn as follows,
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A: Note: As per bartleby guidelines for more than one question asked only one question to be solved…
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A: Drawing the free body diagram of the given system, From the free body diagram, Angle…
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A: For solution refer below images.
Q: 4.5KM/M 225 mm 8m 125mm 35mm
A: Given Data: The magnitude of load, w=4.5 kN/m
Q: D SUMMARY OF ANWERS: МЕMBER CROSS-SECTIONAL AREA, IN² BD 12 ft 8 ft |c 8 ft 36 kips BE E CE 8 ft 8…
A: Determining reactions at support, Consider vertical equilibrium,…
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A: A beam is defined as a structural member subjected to transverse shear load during its…
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A: Since on the system, some of applied forces are acting at some angle, so it is better to resolve…
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A: Drawing the system of force as shown below – Let us take point A as an origin of the coordinate…
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Q: 20 KN 40 kN/m 150 kN m 3m
A: According to the company's guidelines when multiple questions are asked in a single request, only…
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A: given; upper left beam; ⇒lets take reaction force at A=RA⇒lets take reaction force at B=RB⇒load…
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- The 40-kghomogeneous disk is placed on a frictionless inclined surface and held in equilibrium by the horizontal force P and a couple C (C is not shown on the figure). Find P and C.The 40-lb spool is suspended from the hanger GA and rests against a vertical wall. The center of gravity of the spool is at G and the weight of the hanger is negligible. The wire wound around the hub of the spool is extracted by pulling its end with the force P. If the coefficient of static friction between the spool and the wall is 0.25, determine the smallest P that will extract the wire.The coefficient of rolling resistance between the 30-kg lawn roller and the ground is r=0.1. (a) Determine the force P required to pull the roller at a constant speed. (b) What force P would be needed to push the roller at a constant speed?
- The contact surface between the 36-lb block and 20-lb homogenous cylinder is frictionless. Can the system be in static equilibrium 0n the rough inclined plane?The spring attached to the homogenous bar of weight W is undeformed when =0. Determine the smallest spring stiffness k for which the =0 equilibrium position will be stable. Use W = 10 lb, a = 24 in., and b = 6 in. Assume that the spring remains horizontal, which is a valid approximation if is small.The 14-kN weight is suspended from a small pulley that is free to roll on the cable. The length of the cable ABC is 20 m. Determine the horizontal force P that would hold the pulley in equilibrium in the position x=5m.
- Determine the magnitude and direction of the angle theta of the force FAB that must be exerted on bar AB to maintain equilibrium of the system. The suspended mass is 110 kg. Neglect the size of the pulley at A and treat all pulleys as smooth. What should happen to maintain the equilibrium of the system if FAB acts with an angle theta= 0°?Consider g = 9.81 m/s2.the system blocks shown below is in equilibrium, the weight of block A is 10 kN and for block B is 35 kN . The coefficient of friction between A and surface is 0.2 and between two block A and B is 0.15. Answer the following: The direction of friction force between two blocks which effect on block A is in * First Quarter Second Quarter Third Quarter Fourth Quarter The value of friction force between two blocks is * 5.6 kN 6.4 kN 4.8 kN Non above-mentioned The value of friction force between block A and its surface is * 4.0 kN 5.5 kN 6.2 kN Non above-mentioned The coefficient of friction between block B and surface is * 5 نقاط 0.24 0.63 0.56 0.37 The value of friction force between the block B and it's surface is * 25 kN 35 kN 15 kN Non above-mentioned The system is in * Equilibrium Impending to motion KinematicThe winch cable on a tow truck is subjected to a force of T = 8 kN when the cable is directed at = 55°. If the truck has a total mass of 3.5 Mg and mass center at G, determine the following: A.) The magnitude of the total brake frictional force F for the rear set of wheels B. B.)The magnitude of the total normal forces NA at both front wheels A for equilibrium. C.)The magnitude of the total normal forces NB at both front wheels B for equilibrium.
- If d=2m and F-200N and the mass of ball 30 kg Determine the forces in cables AC and AB in equilibrium 1.5 m F -2 m- Maximum size for new files: 128MBHello good evening, Permission, i have a question in my homework. The following bellow is question. Please advice. Thank you Regards,Irfan A 50 kg cabinet is mounted on casters that can be locked to preventtheir rotation. The coefficient of static friction between the floor and each caster is 0.30 and h = 80 cm. Assuming that the casters at both A and B are locked, determine :(a) the force P required to move the cabinet to the right,(b) the largest allowable value of h if the cabinet is not to tip over.An athlete of 60 kg and 1.70 m tall performs the ring exercise called "the Christ", in which he keeps his body immobile with his arms extended horizontally as shown in Fig.-Prob.17(a). The angle with the vertical of the cords from which the rings hang is theta = 10. (a) Determine the tension T in the cords holding the rings. (b) Considering each arm of the athlete as a rigid horizontal bar subjected to the forces indicated in Fig.-Prob.17(b), calculate the value of the components RX and RY, the value of the reaction “R” and of the beta angle.- W is the weight of the arm, applied in the middle of its length.- R is the reaction at the shoulder joint O.- RX and RY are the horizontal and vertical components respectively of the reactionapplied at the shoulder joint O.- Note: Consider that the mass of each arm of the athlete is 3% of the total mass, and thatthe length of the arm is equal to 35% of its height. Fig.-Prob.17(a): Man performing the position of "The Christ"Fig.-Prob.17(b):…