1)a man uses a rope to haul a load of 750n up an wooden inclined plane of effective length 4.50m,1.5m high,the fractional force between the plane and the is 200n.find the;
A)effort required to move to haul up the plane.
B)volocity ratio
C)useful work done on the plane.
2) a four pulley system with a continous rope round it.supports a load of 600n.the efficiency of the system is 75%,if an effort raises the steadily throung a vertical distance of 2m.
Calculate;
a) effort
B) workdone by the effort
C)work wasted.
3) a screw jack whose pitch is 2mm is used to raise a car of mass 900kg throung a height of 20cm.the length of the tommy bar of the jack is 40cm,if the jack is 60% efficient.
Calculate the;
A) volocity ratio
B)mechanical advantage of the jack
C)effort required
D)workdone by the effort.take g=10m,pi=22/7.
4) an inclined plane of an angle 30 is used to raise a load of 4500n through a height of 2m.if the plane is 75% efficient.
Calculate the;
A)mechanical advantage
B)volocity ratio
C) workdone by the effort.?
I already took the course and do not need practice. Please let us see your attempts so we know where you got stuck.
(Velocity by the way, and if you just pull it with a single rope and no pulleys,it moves just as fast as you do. )
Questions
1) To solve this problem, we need to use the concept of mechanical advantage, which is the ratio of the load to the effort.
A) The effort required to haul up the plane can be calculated using the equation:
Effort = (Load / Mechanical Advantage) = (750N / (200N / 750N)) = 750N
B) The velocity ratio 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 effective length of the inclined plane (4.50m), and the distance moved by the load is the height of the inclined plane (1.5m). Thus, the velocity ratio is:
Velocity Ratio = Distance moved by effort / Distance moved by load = 4.50m / 1.5m = 3
C) The useful work done on the plane can be calculated using the equation:
Useful Work = Effort x Load = 750N x 750N = 562,500 Joules
2) In a four pulley system, each pulley increases the mechanical advantage by a factor of 2.
a) The effort can be calculated using the equation:
Effort = Load / Mechanical Advantage = 600N / 2^4 = 600N / 16 = 37.5N
b) The work done by the effort can be calculated using the equation:
Work Done = Effort x Distance = 37.5N x 2m = 75 Joules
c) The work wasted can be calculated using the equation:
Work Wasted = Work Done - Useful Work = 75 Joules - (600N x 2m x 0.75) = 75 Joules - 900 Joules = -825 Joules. Since the work wasted is negative, it means that the system is not 100% efficient and some work is lost.
3) To solve this problem, we need to use the formula for mechanical advantage in a screw jack, which is the ratio of the pitch of the screw to the length of the tommy bar.
A) The velocity ratio can be calculated using the formula:
Velocity Ratio = Distance moved by effort / Distance moved by load = Pitch / Length of the tommy bar = 2mm / 40cm = 0.05
B) The mechanical advantage can be calculated using the formula:
Mechanical Advantage = Velocity Ratio x Efficiency = 0.05 x 0.60 = 0.03
C) The effort required can be calculated using the formula:
Effort = Load / Mechanical Advantage = 900kg x 10m/s^2 / 0.03 = 300,000N
D) The work done by the effort can be calculated using the formula:
Work Done = Effort x Distance = 300,000N x 20cm = 6,000 Joules
4) To solve this problem, we need to use the concept of efficiency, which is the ratio of the useful work done to the work input.
A) The mechanical advantage can be calculated using the formula:
Mechanical Advantage = Load / Effort = 4500N / (4500N / 2m) = 2
B) The velocity ratio can be calculated using the formula:
Velocity Ratio = Distance moved by load / Distance moved by effort = 2m / 2m = 1
C) The work done by the effort can be calculated using the formula:
Work Done = Load x Height = 4500N x 2m = 9000 Joules
Note: The work input is not provided, so we cannot calculate the efficiency or the work wasted.