00-0.38m 680N 0.19m motor exerts up to 1854m (6) What is the maximum possible angular acceleration when lifting a 680N load?

PART 3. VELOCITY ANALYSIS OF ROTATING CRANKS. INSTRUCTIONS: ON A SEPARATE PIECE OF BOND PAPER. SHOW THE COMPLETE SOLUTION AND FINAL ANSWER OF THE FOLLOWING PROBLEMS. 3,600° 1. The instantaneous angular velocity of the crank Q2A is counterclockwise. Q2A = 1.5 in. Find the second instantaneous linear velocity. And draw a graphical interpretation showing the proper length and direction of the velocity arrow of Q2A. Velocity scale: 0.5 in = 10 in/s. Drawing scale: 1:1 2.50 1.50 60° 60°

Theory of Machines -Thired Stage. a/ A pinion having 20 in volute teeth of module pitch 6mm ayear wheel rotutes at 2o r.p.m. and traurmits /.5kw to l having 5o teeth The adden dum on both the wheels is Vy of the circular pitch -The angle of obliquity is 20' Find ca) the bength of the path of appsoach (b) the length of the arc of approach,(c) the normal force betmen the teeth atn Instant where thare is only pairof tath in Con tact. [Ans. 13.2Fmm, 14.02 mm. il 93NJ

Link AB has a clockwise angular velocity waR-61 rad/s and and increases at a rate of aAR-60 rad/s 90 mm D 90 mm В 90 mm O A 216 mm -216 mm

A wheel rest at the corner of the block as the figure. It F is a minimum force required to raise the wheel onto the block, What is the force at Point A ? The block is fixed

1. Calculation for slider crank When, ? = 100 rad/???2 R=62.5mm L=260mm ? = 90 ? = 100 ???/??? Find : 1-find displacement(x) 2-find velocity (v) 3-find acceleration (?)

Problem 6: The angular velocity and angular acceleration of bar AB are @AB = 2 rad/s and aAB =.8 rad/s". What is the acceleration of point D? 0.32 m 0.48 m Answer: an = -0.697i - 1.71j rad/s. A AR -0.32 m--0.24 m-0.16 m- cpng

The figure below shows the components of normal forces, the value of radial force on the gear wheel due to a tangential force of 636 N is N, Take pressure angle as 14½°. ....... af PN Pitch Pt point Gear driven 02 The radial force is

1-)The diagram of the change of cutting force for the cutting work performed with the press seen in the figure is given in the figure Engine revthe number is 900 rpm and the number of Flywheel revolutions is 60 rpm.The moment of inertia of the rotating masses in the engine is 0.35 kgm^2. Speedcalculate the moment of Flywheel inertia required for the press so that its smoothness is 0.01 . Engine power is 1.5 kWit is known

A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10.0 s period following the inital spin, the bike wheel undergoes 45.0 complete rotations. Assuming the frictional torque remains constant, how much more time At, will it take the bike wheel to come to a complete stop? Ats = The bike wheel has a mass of 0.725 kg and a radius of 0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque Tf that was acting on the spinning wheel. N. m Tf = - TOOLS x10