A jackscrew has a lever arm of 40cm and a pitch of 5mm .if the efficiency of the machine is 20 percent ,what force F is required to lift a load w of 280kg​

Respuesta :

Answer:

Theoretical M.A. = 40 cm / .5 cm = 80

Actual M.A. = 80 * .2 = 16

F * 16 = 280 kg       F here is actually mass because 280 kg is mass

F = 280 kg / 16  = 17.5 kg

F (Newton's) = 9.8 m/s^2 * 17.5 kg = 172 N

27.5N of effort force is required to lift a load w of 280kg​

Define effort force ?

The force used to move an object over a distance. resistance force: The force which an effort force must overcome in order to do work on an object via a simple machine.

given

length of liver arm = 40 cm = 0.40 m

pitch = 5mm = 0.005 m

IMA = 2[tex]\pi[/tex](0.40) / pitch

      = (2 * 3.14 * 0.40) / 0.005

      = 0.50 * [tex]10^{3}[/tex]

since , efficiency = AMA / IMA

AMA = 0.20 * 0.50 * [tex]10^{3}[/tex]

         = 0.1 *[tex]10^{3}[/tex]

AMA = Force of resistance / effort force

0.1 *[tex]10^{3}[/tex]   =  280 * 9.81 / effort force

effort force =  280 * 9.81 /0.1 *[tex]10^{3}[/tex]  = 27.47 N≈ 27.5N

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