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- Find the instantaneous velocity of slider F, in mm/s using relative velocity method. The crank AB rotates in uniform angular speed of 150 rpm in counter clockwise direction. AB = 50 mm, BD = 400 mm, DC = 150 mm, CE = 100 mm, EF = 95 mm, DC and CE are perpendicular. Draw in appropriate scale. USING THE GRAPHICAL APPROACH (AutoCad)rad The rod OA rotates clockwise at a constant angular velocity of 0 = (3) Two pin-connected slider block, located at B, move freely on OA. The curved rod is described by the equation r (300 (3-cos)) m. Hint: 1. Remember clockwise rotation is negative velocity. OA Variable 0₁ d₁ d₂ d3 0₁ d2 d3 Values for the figure are given in the following table. Note the figure may not be to scale. Value 110 degrees 0.45 m 0.1 m 0.275 m Using cylindrical components, a. Determine the cylinders's radial and transverse components of velocity at the instant shown, u, and vg. b. Determine the cylinders's radial and transverse components of acceleration at the instant shown, ar and ag. c. Determine the cylinder's magnitude of the velocity at the instant shown, ·M. d. Determine the cylinder's magnitude of the acceleration at the instant shown, |a1. Round your final answers to 3 significant digits/figures. 7= fr + a= M-C Hr+ m S M 82 He) Fe) m S mA wheel rotating with 17 rev/s is uniformly accelerated at 0.5 rad/s² for 5 seconds. It then rotates with a uniform speed for next 7 s. Then it is made to stop in next 15 s. Determine (i)maximum angular velocity; (ii) total angular displacement (Enter only the values in the boxes by referring the units given . Also upload the hand written copy in the link provided) Solution: The maximum velocty in rad/s is The total displacement in rad is
- Show full solution pleaseFind the instantaneous velocity of slider F, in mm/s. The crank AB rotates inuniformangular speed of 150 rpm in counter clockwise direction. AB = 50 mm, BD = 400 mm, DC=150 mm, CE = 100 mm, EF = 95 mm, DC and CE are perpendicular. Drawin appropriate scale.(CLUE ANS: 341.6 mm/s)Given the image below, if AB is 0.5m lopng, Rod BD is 1.5m long, and the velocity of point B is 0.5m/s ↖45.0°, what is the speed of piston at D? choices: a. 0.5 m/s b. 0.354m/s c. 1.061 m/s d. 1.5m/sGear A of the winch turns gear B, raising the hook H. Gear A starts from rest at t=0 and its clockwise angular acceleration (in rad/s) is given as a function of time by 0=0.28t. What is the upward velocity of the hook and what vertical distance has it risen at t=12s ? Figure Q5 - Gear winch H
- The figure shows a variation of the Scotch-yoke mechanism. It is driven by crank 2 at an angular velocity and acceleration of 36 rad/s ccw and 120 rad/s2 ccw, respectively. Find the velocity and acceleration of the crosshead, link 4.|AO2|=75 mm. (Using Complex Algebraic approach do the Position Analysis; Velocity Analysis and Acceleration Analysis then solve the problem please)The figure below depicts the linkages of a pump mechanism. If Q2A is rotating at 10 rpm in a clockwise direction, determine the linear velocity of slider D when Q2A has moved 60 degrees from its original position. Use R&C method. (Ks: 1 in. = 1 in.; Ky: 6 fpm = 1 in.) LINEAR VELOCITY OF B = EPM LINEAR VELOCITY OF C = FPM LINEAR VELOCITY OF D = FPMThe acceleration of A is 3.3 m/s? dovwnward and the acceleration of Cis 2.5 m/s? upward. Find the acceleration of B. The acceleration of Bis positive if downward, negative if upward. B A. Part 1 Write an expression for the length of the rope. Sc B A Answer: L= SA+ Sc+ constants Save for Later Attempts: 0 of 1 used Submlt Answer
- For the Mechanism shown, find the acceleration of the slider link4 using Vector loop method. Assume r3=1 cm. Y B KE 13 =90°. 00₂= 10 rad/s CCW (constant) X Hand written plz aQ1. Member AB is rotating (about fixed point A) with angular velocity (@ = 1 rad/sec) and angular acceleration (a = 2 rad/sec?) and driving the link CD (rotating about fixed point D at constant velocity) as shown in figure. Select five right answers. (a) Tangential Acceleration (a:) of link AB will not be zero (b) Tangential Acceleration (a) of link AB will be zero D 450mm (c) Tangential Acceleration (a,) of link CD will not be zero 200 mm 100 mm 60 H (d) Tangential Acceleration (a,) of link CD will be zero (e) All points on link CB will have same linear velocity 0 All points on link CB will have same angular velocity Formula • Tangential acceleration of a rotating link -ra • Radial (normal) acceleration of a rotating link = v³ /r= wr velocity of a rotating Ink (v) = r.w (g) Angular acceleration (a) for link CD will be zero (h) Angular acceleration (a) for link CB will not be zero Q2. Magnitude of linear velocity of any point “H" located at the half of the length of a rotating link CB…Draw a six-bar linkage mechanism used for holding an industrial stirrer. Points A, B, and E are on a straight line on link 2. Points B, C, and G are on a straight line on link 3. The input link 2 rotates with a constant angular velocity a of 3 rad/s (cw) at the instant. O2=0 (a) Complete the acceleration polygon according to the drawing scale given, and find the angular accelerations Oy, 04, as, and as of links 3, 4, 5, and 6, and the acceleration Ap of point F. The position of link 2 is given as input. Write down the vector loop equations of the linkage for analytical position analysis and identify the unknowns in the equations. Note: Round off all your answers to two decimal places. Dimensions: AB= 40.00 mm BE = 20.00 mm BC = 33.54 mm CG= 33.54 mm CD = 18.03 mm EF = 43.00 mm FG = 29.15 mm