A screw jack has 90 threads tothe metre. the effort is applied at the end of an arm 20cm long whatforce must be applied to lift a load of 150N ?
A screw-Jack has 90 threads to the metre, the effort is applied at the end of the aim 20cm long what force must be applied to lift a load of 150N
133
how do you arrive at 133?
Well, to calculate the force needed, we need to use the formula for the mechanical advantage of a screw jack. But hold on, let me grab my tiny screwdriver and clown-sized calculator. Ah, here it is!
So, we know that the screw jack has 90 threads per meter. Therefore, for every 1 revolution, the screw jack will move 1/90th of a meter. Now, let's convert the length of the arm to meters. That's 20/100 = 0.2 meters.
To find the mechanical advantage, divide the distance moved by the screw jack by the distance applied by the effort. In this case, the screw jack has to move 0.2 meters to lift the load 150N. Dividing these two, we get 0.2/0.00111 (1/90th of a meter) = around 180.
So, the mechanical advantage is approximately 180.
Now, to find the force needed, divide the load (150N) by the mechanical advantage (180). That's 150/180 ≈ 0.8333333333333333 (or approximately 0.83N).
So, you would need to apply a force of around 0.83 Newtons to lift that 150N load. Just make sure to stretch your funny bone a bit while turning that screw jack!
To calculate the force required to lift a load using a screw jack, we need to use the formula:
Force (F) = (Load * Distance)/Effort Arm Length
In this case, the load is given as 150N, and the effort arm length is given as 20cm. However, we still need to determine the Effort (E), which is the force applied at the end of the arm.
The Screw Jack has 90 threads to the meter, which means that for every full turn of the jack, it moves 1/90th of a meter.
To calculate the Effort (E), we need to convert the effort arm length from centimeters to meters. There are 100 centimeters in a meter, so the effort arm length is 0.20 meters.
Now we can calculate the Effort (E):
E = (F * d) / L
Where:
E = Effort
F = Load (150N)
d = Distance covered by 1 full turn of the screw jack (1/90th of a meter)
L = Effort arm length (0.20 meters)
E = (150N * (1/90 m)) / 0.20 m
Simplifying the equation:
E = (150N / 90) / 0.20
E = 1.67N
Therefore, the force that must be applied to lift a load of 150N with a screw jack is approximately 1.67N.
each turn lifts 100/90 cm or 1.11 cm. MA is distanceup/distancearmmoves in one rev
force= 150(1.11/20PI)=2.65 N