Determine the maximum weight of the engine that can be supported without exceeding a tension of 450 lb in chain AB and 480 lb in chain AC with 0 = 30.0°. B а) Draw the free-body diagram for point A below Co000
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- Find the stable equilibrium position of the system described in Prob. 10.56 if m = 2.06 kg.Equilibrium of Concurrent Force System. The piston of a reciprocating engine exerts a force of 175 kN on the crosshead when the crank is 45° (angle B) past TDC. If the stroke of the piston is 800 mm and the length of the connecting rod is 1.80 m, find the guide force and the force in the connecting rod. Hint: 1=½ stroke; Angle o may be solved using Sine Law in AACO. K= Compressive Force in the connecting Rod E= Piston Effort 70 120 F, = Guide Force Fig. P-1 Fig. P-2 a Tools: EF cos0=0; F sin .=0. R = Ev)* + (E#): ; R 771 =- : tane = sin A sin B sin CThe diameter of two pulleys are (0.3m)and (0.2m), center distance (1m).if we use open belt drive. Determine: 1- the length of V- belt (2ß = 60°). 2- the contact angle between belt and smaller pulley. 3- the horse power, if the smaller pulley turn with speed (400r.p.m) and the tension force in tight side (T1=90ON), coefficient of friction u=0.3 4- the speed of large pulley.
- 2. Find the horizontal force P to start the motion of any part of the system of three blocks resting upon one another as shown. The weight of the blocks are W-300 KN, Ws= 100 KN and Wc =200 KN. Between A and B μ=0.30, between C and B, μ =0.20 and between C and the ground, p=0.10. 1 sni brod nog 02.0-4 B C 05.0-4 Solution: tamos quoy worla bre sosia asquar 005 300 en of jells400 N₁-300 F3 N-300- N₂=300 Assuming B and C will not move. F₁ = 300 (0.30) F₁ = 90 kN ne à bns A zelboß 1 P = 90 kN es ensla berb hebnu 02.0- y elin 05.0= q V in Celpris orllanimetsa MX 008 Assuming C will not move: al 1st F2=400 (0.20) = 80 kN OTE Bm ribirte F₂ vica Assuming the three blocks move: F3 = 0.10 (600) F3 = 60 kN P=60 KN Least force is 60 kN.The 300kn sphere is supported by a pull p and a 200kn weight passing over a frictionless pulley. If a=30°. Computer the value of p and tangent.2. A 4-lb weight stretches a spring 6 in. The weight is pulled 3 in below the equilibrium position and released. Find the motion of the weight if an cos6t acts on the system. external force f(x) = 4 F(x)= Ma -what is the spring constan" K-F 4 7=33
- The car shown in the figure below has a mass of 1673 kg. The coefficient of static friction between the rubber tires and the pavement is 0.6. Determine the maximum incline e [degrees] that the car can drive up if it has rear-wheel drive. 0.85 m 1.2 m 1.7 m Answer: 18.87 Calculate the normal reaction force at the rear wheels for the condition of rear-wheel drive in N. Answer:Pulley A delivers a steady torque of 1100 lb-in. To a pump through its shaft at C. The tension in the lower side of the belt is 165 lb. The driving motor weighs 200 lb and rotates clockwise. Determine the bearing reaction force at O. A 9" B 30° 37 8" 150 lb - 5"-- 5"-The figure below shows a cylinder, which is rotating in clockwise direction. A lever is pushing onto the cylinder, thus exerting a braking torque. The lever itself is loaded by a force FE . a) Draw the free-body diagram of the lever, determine the direction of the kinetic friction force.
- Question 2: The ideal spring of constant k-2.6 kN/m is attached to the disk at point A and the end fitting at point B, as shown. The spring is unstretched when OA and Oв are both zero. If the disk is rotated 15° clockwise and the end fitting is rotated 30°counterclockwise, determine the vector expression for the spring force F. ( Determine distance C so that the moment the spring force makes about the Z axis is equal to 10.82 N.m. ( A = 15°. C A 250 mm eeeeeee 900 mm k= 2.6 kN/m OB = 30° B LC 7-y 200 mm3/13 Three cables are joined at the junction ring C. De- termine the tensions in cables AC and BC caused by the weight of the 30-kg cylinder. 45° D C 15° 30 kg 30 Problem 3/131. In lecture I focused on a horizontal mass-spring system so we could ignore gravity, but in practice it's much easier to build a vertical mass-spring system. In the figure below, we can see a suspended spring with spring constant k both before and after a mass m is hooked to it. Y Yo yo is the equilibrium height of the spring (without the mass) and in the figure y = 0 is set at the ground. a) Using the coordinate system in the figure, show that Newton's second law for the hanging mass-spring system leads to the following differential equation: d²y dt² k (y - Yo) - g m b) Now define a new coordinate, y', related to y by a constant shift: y = y' + C, where C = constant. Show that if you choose the right value of C, Newton's second law is identical in form to a mass-spring system without gravity. What is the period of the system? c) Effectively, gravity just shifts the equilibrium position of the system, but the system still undergoes simple harmonic motion. What is the net force on the…