Determine the Grashof condition and Barker Classification for the following mechanism: 20 cm 40 cm 30° 18 cm The value S+L= The value P+Q= The Grashof Condition is: The barker classification is: 4 35 cm
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- ....A steam engine develops power at mean torque of 19,000 N-m and velocity of 10 m/s. The mean diameter of the flywheel rim is 1.2 meters. The coefficient of fluctuation of speed is 0.03 and the coefficient of fluctuation of energy is 0.21. Determine the thickness of the rim in mm if the width of the rim is twice of its thickness. Assume density of rim material as 7250 kg/m3.The controlling force curve of a spring controlled governor is a straight line. The weight of each governor ball is 40 N and extreme radii of rotation are 10 cm and 17.5 cm. If the values of the controlling force at above radli are 205 N and 400 N respectively and friction of mechanism is equivalent to 2.5 N at each ball. Find The extreme equilibrium speeds of the governor. (1) The equilibrium speed and coefficienct of insensitiveness at a radius of 15 cmQ4.A two-degree-of-freedom model consisting of two masses connected in series by two springs is shown in the figure below. The physical parameters have the values ?1 = 8 kg, ?2 = 2 kg, ?1 = 20 N/m, and ?2 = 30 N/m. (C) Calculate the first (larger) natural frequency of the system (D) Calculate the second (smaller) natural frequency of the system
- Determine the natural frequency of the spring-mass system as shown in Figure. The values of spring constants K1=K2=K3=K4=K5=K6=100 N/m. The mass m=10kg.Required to answer. Single choice. OPTIONS: 1.About 1.211 Hz 2.About 0.871 Hz 3.About 0.520 Hz 4.About 0.620 Hz 5.NoneDimensions and dimensions in the system given in the figure: L1 = 850 mm; L2 = 1700 mm; L3 = 2400 mm; U4 = 3100 mm m1 = 2 kg; m2 = 3 kg; m3 = 4 kg R1 = 1135 mm; R2 = 820 mm; R3 = 1040 mm θ1 = 113o; 2 = 50o; θ1 = 250o Find in mA, RA, θA and mB, RB, θB required to achieve dynamic balance?The equivalent spring stiffness of an inclined spring whose k = 5 N/m and angle of inclination θ = 35° 3.355 N/m 4.095 N/m 5.867 N/m 4.511 N/m
- In the system shown,m= 2 kg, k= 1000 N/m, theta= 30 degrees and the friction coefficient between the block and the surface is u. (a) If the block is displaced from equilibrium and released, the decrease in amplitude per cycle is 2 mm. Determine the coefficient of friction u. (b) If the initial displacement is 5 cm from equilibrium, determine the time and position when the block comes to rest.Can anyone help me with this question. The Motor (J1=0.46 kg‐m2) is connected to a gearbox (J2 =0.27 kg‐m2 and J3=0.17 kg‐m2) by a steel shaft. The output of the gearbox is connected to a turbine (J4=0.37 kg‐m2 by a second steel shaft. The shafts are supported by bearings (not shown). The gearing in the gearbox is such that theta3=‐2.7theta2. Use Lagrange methods to determine the equation of motion for the system including the flexibility of the shafts, using variables of theta 1, theta 3, theta 4 based on changing theta 3.A spring-mass system is shown in the following figure. Use m=3.5 kg and k=100 N/m. a. Determine the static deflection δst . Insert your answer in meters correct up to at least a third decimal place. b. Determine the system period τ. Insert your answer in seconds correct up to at least a third decimal place. c. Determine the maximum velocity vmax which results if the cylinder is displaced 140mm downward from its equilibrium position and released from rest. Insert your answer in m/s correct up to at least a third decimal place.
- In the system shown in the figure, the spring stiffness k = 2 × 103 N/m and the total massof the system is m = 20 kg, the eccentric mass m0 = 0.2 kg, the eccentricity e = 0.1 m andthe rotation angular velocity of the unbalanced mass is ω1 = 20 rad/s. The applied force isF(t) = 15 sin ω2t with ω2 = 30 rad/s. Determine:(a) the steady state motion of the system;(b) the force transmitted in the spring and estimate the maximum value of the forceThe turning moment diagram for a multi cylinder engine has been drawn toa scale of 1 mm = 4500 N -m vertically and 1 mm = 2. 4 ͦ horizontally. Theintercepted areas between output torque curve and mean resistance lineabove and below the mean taken in order from one end are 342, 23, 245,303, 115, 232, 227, 164 ??2, when the engine is running at 150 rpm. Ifthe mass of the flywheel is 1000 kg and the total fluctuation of speed doesnot exceed 3% of the mean speed, find the minimum value of the radius ofgyration.The figure shown below is released from rest with the spring unstretched. Use the datum shown for zero gravitational potential energy. For the values given below answer the following questions:Position 1 is Ha1 μk=0.14 k=32.5 Nm hA1=0.6 m mA=4.6 kg mB=1.3 kg What is the Gravitational Potential Energy for mass mA at position 1 (remember your signs)? What is the Gravitational Potential at position 1 for mB? What is the Potential Energy of the spring at position 1? What is the Kinetic Energy at position 1 for mA? What is the Kinetic Energy at position 1 for mB? Again we have the figure shown below that was released from rest with the spring unstretched. Use the datum shown for zero gravitational potential energy. At position 2 the spring is at its max stretch. Remember we have the values below; answer the following questions: μk=0.14 k=32.5 Nm hA1=0.6 m mA=4.6 kg mB=1.3 kg What is the max stretch in the spring?