Question

Explain to the smallest details the solution done below.

5/153 The clemcnts of a power hacksaw arc shown in the figure. The saw blade is mounted in a frame which slides along the horizontal guide. If the motor turns the flywheel at a constant counterclockwise speed of 60 rev/min, determine the acceleration of the blade for the position where θ-90", and find the corre- sponding angular acceleration of the link AB. 100 mmm 450 mm 100 mm Problem 5/153 Reference system: horizontal x-axis with positive orientation to the right; vertical y-axis with positive orientation upwards. Angular velocities and accelerations are considered positive when CCW 0 0 0 60 ω0B2 :___ . 2.7-6.283 60 (because the angular velocity is constant) float,4[o 6.283 0 0 Position vectors (when θ-90 deg) -0.1 -0.439 0 0 OB: Fixed axis rotation 2float,43.948 0 AB: General motion oBA0 -0.143 4.894 1.432 0.467 float,4 4.894 (Acceleration of the blade) 0 float,4 (Angular acceleration of link AB)

Answer 1

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First we take a reference axis system centered at point '0' as shown belocw Note that z-axis is perpendicwar ^ to the paper about which the motor turns. Counte-clo ckwise ccw rotations are taken as positive. R100 i00mm 21-60=6.283 Tad/s (CCW 60 60 (6283 k) Tad/s ce OB since the motor is rotating at constant angular vel ocity Position vectos BDIS dCAO-Jo.452-0.12= 0.439 m d(AD) 0.439+ 0.1 0.439+o.1- o. 539 m YAB = 8-A = (-0.1-(-0, 539)) 1 + (0-(-0.1〉J + (o-W k TAB 0.439 L +0.1j

We have VBE ώ0BXYBo -(6. 283 k)x (o.1 i and, 2→ We drao velocty tiangle ve measure VA 0.143 í) mis VAB-0.6d4 rols VaB oBA- VBA 0.Gu1. 432 radls O.4S (0.6283)! YBA COBAニ(-1. 432 k) Tad's Noo we draw acceleration Polygon. (ar)AB- VAB0-0.922 m/s2 we me asure OA = 4, 834 m/s. aA oe 2. AB O.45 3 2. o.210 at) A8-2101 10 O-45