Simple Machines: Understanding Screw Jack Efficiency
How can we calculate the velocity ratio, mechanical advantage, effort required, and work done by the effort of a screw jack?
Given that a screw jack with a pitch of 2mm is used to raise a block of 700kg through a height of 30cm, and the length of the tommy bar of the jack is 50cm, with 70% efficiency.
Calculations:
1) The velocity ratio is \(500\pi\)
2) The mechanical advantage is \(350 \pi\)
3) The effort required is 6.2 N
4) The work done by the effort is 2922 J
In solving for the various parameters of the screw jack, we start by calculating the velocity ratio, mechanical advantage, effort required, and work done by the effort:
Velocity Ratio Calculation:
The velocity ratio of a screw jack is determined by the equation \(VR = \frac{2\pi L}{p}\), where \(L\) represents the length of the tommy bar and \(p\) is the pitch.
Given that the length of the tommy bar is 50cm and the pitch is 2mm, we can substitute these values into the equation to find the velocity ratio as \(VR = \frac{2\pi(50)}{0.2} = 500\pi\).
Mechanical Advantage Calculation:
The mechanical advantage of a screw jack can be calculated using the formula \(MA = \frac{Load}{Effort}\), where the load is the force exerted on the output side.
For this scenario, the load is the weight of the block lifted by the screw jack, which is 700kg. By substituting the values into the equation, we can determine the mechanical advantage as \(MA = \frac{700 \times 9.8}{350\pi} = 350\pi\).
Effort Required Calculation:
The effort required to operate the screw jack can be found using the formula \(Effort = \frac{Load}{MA}\).
Substituting the known values into the equation, we can calculate the effort required as \(Effort = \frac{6860}{350\pi} = 6.2 N\).
Work Done Calculation:
The work done by the effort on the load can be determined by the equation \(W = (Effort) \times VR \times d_{load}\).
Given the values of effort, velocity ratio, and the distance of the load, we can substitute them into the formula to find the work done as \(W = (6.2) \times (500\pi) \times (0.30) = 2922 J\).
By understanding and applying these calculations, we can gain insights into the efficiency and effectiveness of using a screw jack in lifting heavy loads.