s***w jack - velocity ratio/mechanical advantage?

Question

A s***w jack has a pitch of 4mm. the lever used to turn the s***w is 150mm long. the efficiency is estimated at 25%. find:

a)the velocity ratio

b)the mechanical advantage

c)how much force is required at the lever to lift a mass of 300kg?

take g = 9.8 m/s^2

the question is simple but i keep getting wrong answers !! please help !!

Answer 1

Read below and judge your-self if there is anything wrong.

W = weight

F = smaller force  applied at the handle.

R represents the length of the handle and

P the pitch of the s***w, or the distance advances in one complete turn.

F*2*PI*R = W*P

W=M.g

W=300*9.8

F= [W*P] / 2*PI*R  [neglecting friction]

F=300*9.8*[4/1000] / 2*3.14*[150/1000]

F=11.76 / 0.942

F=12.48 N

Efficiency=M.A. / V.R.

V.R =M.A../ 0.25

A lifting machine is any machine designed to enable a load (FL) to be raised by a much smaller effort (FE).

The ratio is called the Mechanical Advantage or Force Ratio.

M.A. = FL/FE

M.A.= 300*9.8/12.48

M.A.=235.57

V.R =M.A../ 0.25

V.R =942.28

Answer 2

Not too sure about this one...  So, every time the end of my 0.150m long handle completes a full revolution (the end of the handle travels through 0.150m*2*pi = 0.942m), the jack lifts 0.004m, and the efficiency is 25%?

(a)

vel_end_of_handle/vel_screw = 0.942/0.004 = 236

vel_screw/vel_end_of_handle = 0.004/0.942 = 0.00424

(b)

mechanical advantage = 0.942/0.004 = 236

(c)

0.25*F = 300kg * 9.8 m/s^2 / 236

F = 49.9 N