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2 years ago

Physics law of machine

1 answer:

3 0

1. The problem statement, all variables and given/known data A screw jack has a single start thread of pitch 7mm and a operating handle 800mm long.when raising a load of 750kg the effort required on thev end of the handle is 26N. determine for these operating conditions the following : (a) the mechanical advantage (b) the velocity ratio (c) the efficiency of the machine (d) the law of the machine 2. Relevant equations ma = load/effort vr = 2 x pie x r/p effiency of the machine = ma/vr x 100% law of the machine = E = aw+b 3. The attempt at a solution (a) mechanical advantage = 750 x 9.81/26N = 282.98 (b) the velocity ratio = 2 x pie x 800mm/7mm = 718.07 (c) the effiency of the machine = 282.98/718.07 x 100% = 39.40% (d) the law of the machine this is where i am struggling i no the formula is E = aw+b where a is the velocity ratio and w is the load however what does the b stand for and if i need to caculate this how do i do this please help ......

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Answer:

$16.25^{\circ}$

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