
#1
Jun2108, 04:24 AM

P: 54

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 ...... 



#2
Jun2108, 06:01 AM

P: 200

Substitute the values of E,a,w in the law of machines to obtain the value of b (it is a constant).




#3
Jun2108, 09:11 AM

P: 54

so i have transposed correctly i think and have came up with this :
b = Ea/w therefore b = 26n  718.07/750 x 9.81 = 16.60 thanks for the quick reply 



#4
Jun2108, 02:57 PM

P: 54

The law of the machine
sorry to reply again as i have misread part (e) in the question
it says if the effort needed to raise a load of 400kg is 17N determine : the law of the machine. also should the law of the machine when calculated total 17N thanks again mark 



#5
Jun2208, 12:32 AM

P: 200

However, you have been provided with two sets of values for E and W. Substitute these in the law of machines, and obtain the values of 'a' and 'b' by solving the system of linear equations that you get. 


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