a screw jack whose pitch is 3mm is used to lift a load of 400kg through a length of 14cm,the length bar of jack is 20cm.if the efficiency of the jack is 60% . calculate
a.velocity ratio of the jack
b.mechanical advantage of the jack
c.effort required to lift the load
d.workdone by the effort?
every turn your hand moves around circle with radius 0.20 meters
2 pi r = 1.23 meters
that raised the load 0.003 meters
ratio = 1.23 / 0.003 = 409
that is also the mechanical advantage if efficiency is perfect 100%
but in reality .6 * 409 = 245
400 kg * 9.8 = 3920 Newtons up on load
your hand pushes with 3920 /245 = 16 Newtons hand push
work done on load = 400*9.8 * 0.14 Joules
so work done by hand = 400 * 9.8 * 0.14 / .6 Joules
What is the ratio if the pitch of the screw is 0.8cm and the effort applied at the end of the spanner describes a circle of radius 10cm?
Desame
To solve this problem, we need to understand the concepts of pitch, velocity ratio, mechanical advantage, effort, and work done. Let's go through each part of the problem step by step.
a) Velocity Ratio of the Jack:
The Velocity Ratio of a machine, such as a screw jack, is the ratio of the distance moved by the effort to the distance moved by the load. In this case, the distance moved by the effort is the length of the bar of the jack (20 cm) and the distance moved by the load is the length through which the load is lifted (14 cm).
Velocity Ratio = Distance moved by the effort / Distance moved by the load
Velocity Ratio = 20 cm / 14 cm
Velocity Ratio = 1.43
b) Mechanical Advantage of the Jack:
Mechanical Advantage is defined as the ratio of the load to the effort.
Mechanical Advantage = Load / Effort
Given that the load is 400 kg, we need to first convert it to Newtons (N). Recall that 1 kg is approximately equal to 9.81 N due to the acceleration due to gravity.
Load = 400 kg * 9.81 N/kg
Load = 3924 N
Efficiency is defined as the ratio of the useful work output to the work input. Given that the efficiency is 60%, we can determine the work input using the formula:
Efficiency = (Useful Work Output / Work Input) * 100
0.60 = (Useful Work Output / Work Input) * 100
Solving for Useful Work Output:
Useful Work Output = Efficiency * Work Input
Useful Work Output = 0.60 * Load * Distance moved by the load
In this case, the distance moved by the load is 14 cm.
Useful Work Output = 0.60 * 3924 N * 14 cm
Useful Work Output = 32957.6 N·cm
Now that we have the Useful Work Output, we can calculate the Work Input.
Work Input = Useful Work Output / Efficiency
Work Input = 32957.6 N·cm / 0.60
Work Input = 54929.3 N·cm
The Mechanical Advantage of the jack is the ratio of the Load to the Effort:
Mechanical Advantage = Load / Effort
Mechanical Advantage = 3924 N / (Work Input / Distance moved by the effort)
Mechanical Advantage = 3924 N / (54929.3 N·cm / 20 cm)
Mechanical Advantage = 1.43
c) Effort Required to Lift the Load:
The effort required to lift the load can be calculated using the formula:
Effort = Load / Mechanical Advantage
Effort = 3924 N / 1.43
Effort = 2743.4 N
d) Work Done by the Effort:
The work done by the effort is given by the formula:
Work Done = Effort * Distance moved by the effort
Work Done = 2743.4 N * 20 cm
Work Done = 54868 N·cm
Therefore, the answers to the questions are:
a) Velocity Ratio of the jack = 1.43
b) Mechanical Advantage of the jack = 1.43
c) Effort required to lift the load = 2743.4 N
d) Work done by the effort = 54868 N·cm