Screw jack: mechanical advantage, velocity ratio, efficiency, and law of the machine

[series111]

Homework Statement

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

Homework 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

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 ...

 

[Vagrant]

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

 

[series111]

so i have transposed correctly i think and have came up with this :


b = E-a/w

therefore b = 26n - 718.07/750 x 9.81 = 16.60

thanks for the quick reply

 

[series111]

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

 

[Vagrant]

also should the law of the machine when calculated total 17N

I don't understand what you mean by this.


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.