A screw jack whose pitch is 2mm is used to raise 19 motor car of mass 100kg through a height of 20.0cm,the length of the tommy bar of the jack is 40cm.if the jack is 60% efficient,calculate the
1.velocity ratio
2.mechanical advantage of the jack
3.effort required
4.work done by the effort[Take g=10ms ^2]
1. The velocity ratio of the screw jack is calculated using the formula:
Velocity Ratio = Distance moved by effort / Distance moved by load
In this case, the distance moved by the effort is equal to the pitch of the screw, which is 2mm. The distance moved by the load is the height the motor car is raised, which is 20.0cm.
Velocity Ratio = 2mm / 20.0cm
Converting mm to cm, we get:
Velocity Ratio = 0.2cm / 20.0cm = 0.01
So, the velocity ratio of the screw jack is 0.01.
2. The mechanical advantage of the jack is given by the formula:
Mechanical Advantage = Load / Effort
The load in this case is the weight of the motor car, which is 100kg. The effort is the force applied by the tommy bar. The force applied by the tommy bar can be calculated using:
Effort = Load / Mechanical Efficiency
Where Mechanical Efficiency = Efficiency / 100 = 60 / 100 = 0.6
Effort = 100kg / 0.6
Effort = 166.67kg
Therefore, the mechanical advantage of the jack is:
Mechanical Advantage = 100kg / 166.67kg
Mechanical Advantage ≈ 0.6
3. The effort required is the force in Newtons applied by the tommy bar. We can calculate it using the formula:
Effort = Load / Mechanical Advantage
Effort = 100kg / 0.6
Effort = 166.67kg
Therefore, the effort required is 166.67kg, or approximately 1667N.
4. The work done by the effort is given by the formula:
Work = Effort * Distance moved by effort
The distance moved by the effort is the pitch of the screw, which is 2mm. Converting it to meters, we get:
Distance moved by effort = 2mm / 1000 = 0.002m
Work = 1667N * 0.002m
Work = 3.334 Joules