Book E: Mechanics Part 2 Assignment - Answers with Commentary
Question 1
A force of 80 N is applied to the end of a bar 2.0 m long. The bar is pivoted 0.45 m from that end. Calculate the force required to be applied to the other end to maintain balance.
Solution
In order to balance the beam, the torque exerted on each side of the pivot must be the same.
A force of 80 N (\( F_1 \)) is exerted at a distance of 0.45 m (\( d_1 \)). The torque can be calculateed.
\( \tau = F_1 d_1 = 80 \cdot 0.45 = \) 36 Nm.
The balancing force must exert this same torque to balance the beam.
The pivot is 0.45 m from one end of the beam.
The beam is 2 m long, so the distance from the pivot to the balancing forece end of the beam must be 1.55 m (\( 2 - 0.45 \)). This distance will be referred to as \( d_2 \).
The force can be calculated from the torque.
\( \tau = F d \rightarrow F = \frac{\tau}{d} \)
Plugging in the known distances yirlds the answer.
\( F_2 = \frac{\tau}{d_2} = \frac{36}{1.55} = \) 23.22 N (4 sf) #.
Question 2
A block has a mass of 100kg. Calculate the downward force.
Solution
In this question, it is presumed that the downward force is gravity.
The force due to gravity is given below.
\( F = m g \)
where \( F \) is the force, or weight (newtons or N), \( m \) is the mass (kg), and \( g \) is the acceleration due to gravity (m.s-2). The acceleration due to gravity depends on the planet, but for earth, \( g \) = 9.81 m.s -2.
The downward force is now calculated by plugging in the known quantities into \( F = m g \).
\( F = m g = 100 \cdot 9.81 = \) 981 N #.
Question 3
A plank is 4 m long and suspended between 2 saw stools, one at each end. A 145 kg mass sits on the middle of the plank. Calculate the force on each saw stool.
Solution
This question is about moment-balancing (NOTE: moment is another word for torque).
An object will not move if all of the moments exerted on it are balanced.
It is assumed that the saw stools have no moment resistance, in other words the only forces they can exert are upwards forces.
The moment of the 145 kg mass about the left hand end of the stool is calculated below. Note that the force is the weight of the 145 kg mass. The distance is 2 m, since the mass is in the middle of a 4 m long stool.
\( \tau_1 = F d_1 = m g d_1 = 145 \cdot 9.81 \cdot 2 = 1422.25 \cdot 2 = \) 2844.9 Nm.
The moment of the 145 kg mass about the right hand end of the stool is calculated below. Note that the force is the weight of the 145 kg mass. The distance is 2 m, since the mass is in the middle of a 4 m long stool.
\( \tau_2 = F d_2 = m g d_2 = 145 \cdot 9.81 \cdot 2 = \) 2844.9 Nm.
Note that both moments are equal.
We can now calculate the upward forces \( F_1 \) and \( F_2 \).
The d-value used in both cases is the length of the plank. This will be referred to as \( L \).
\( F_1 = \frac{\tau_1}{L} = \frac{2844.9}{4} = \) 711.2 N (4 sf) #.
\( F_2 = \frac{\tau_2}{L} = \frac{2844.9}{4} = \) 711.2 N (4 sf) #.
Note that both forces are identical. This is due to the mass being in the middle of the plank.
Question 4
Give a practical example of a 3rd order lever (one you have seen) and draw its single line equivalent.
Solution
A common example is tongs, particularly tongs that have the fulcrum at one end.
Question 5
Calculate the torque required to drive a pulley that has a diameter of 0.5 m and is lifting a mass of 10 kg.
Solution
The torque in this case is given by the equation
\( \tau = F r \)
where \( \tau \) is the torque (Nm), \( F \) is the force (N), and \( r \) is the radius (m).
The force is simply the weight of the object. With this knowledge, the torque is calculated. Note the 0.5 m diameter is halved to make it a radius.
\( \tau = F r = m g r = 10 \cdot 9.81 \cdot \frac{0.5}{2} = \) 24.53 Nm (4 sf) #.
Question 6
Calculate the energy required to raise 200 kg up a vertical distance of 27 m.
Solution
The energy, or work required to lift a mass parallel to a force is given below.
\( E = m g h \)
where \( E \) is the work done (J), \( m \) is the mass (kg), \( g \) is the acceleration due to gravity (m.s-2), and \( h \) is the height the mass is lifted (m).
The difference between this equation and torque is that torque is always exerted at right angles to the radius. In this sense, a torque does no work. The force in \( E = m g h \) is being exerted parallel to the direction of movement, therefore work is done.
With that known, the work is calculated.
\( E = m g h = 200 \cdot 9.81 \cdot 27 = \) 52970 J (4 sf) #.
NOTE: The work done against gravity is reversible. Allowing the mass to fall 27 m will release the same amount of energy.
Question 7
A turning moment of 18 Nm is required to tighten a nut on a busbar clamp. What is the minimum force that must be applied to a spanner of effective length 250 mm if the nut is not to be over tightened?
Solution
This is a straightforward application of the torque equation.
\( \tau = F r \rightarrow F = \frac{\tau}{r} \)
In this case, \( \tau = \) 18 Nm, and \( r \) = 250 mm, or 0.25 m. The value 0.25 m will be used to keep the units consistent.
\( F = \frac{\tau}{r} = \frac{18}{0.25} = \) 72 N #.
Question 8
A motor has an input power rating of 500W. If it is 70% efficient, calculate the output in horsepower.
Solution
NOTE: It is generally easier to work in watts until the very end, then convert to horsepower. Doing this avoids the need to divide by 746 until the very end.The output power is calculated using the efficiency equation.
\( P_{\mathrm{out}} = \eta P_{\mathrm{in}} \)
where \( P_{\mathrm{out}} \) is the output power, \( \eta \) is the efficiency (normalised to between 0 and 1), and \( P_{\mathrm{in}} \) is the input power. The output power will be in the same units as the input power.
The input power \( P_{\mathrm{in}} \) is 500 W, and the efficiency \( \eta \) is 0.7 (this is the same as 70%).
The output power can be calculated.
\( P_{\mathrm{out}} = \eta P_{\mathrm{in}} = 0.7 \cdot 500 = \) 350 W.
The question is asking for the output in horsepower (hp). 1 hp = 746 W.
\( P_{\mathrm{out}} = \frac{350}{746} = \) 0.4692 hp (4 sf).
Question 9
A pump raises 0.3 cubic metres of water per minute from a well that is 3m deep. Calculate the input power to the pump if it is 80% efficient.
Solution
In order to calculate the input power, the required output power requires calculation.
The pump rate is 0.3 m3.min-1. Given that water has a density of 1000 kg.m-3, the pump is extracting 300 kg.min-1 of water.
The unit of power (watts) is joules per second, so it makes sense to convert the water extraction rate to kg.s-1 for further calculations.
The value of this quantity is 5 kg.s-1. This quantity will be given the symbol \( \dot{m} \). Note the dot over \( m \). This means it is a rate of mass transfer.
The equation for the output power given a constant rate of mass transfer is given below.
\( P = \dot{m} g h \)
where \( P \) is the rate of work being done (J.s-1 or W), \( \dot{m} \) is the mass transfer rate (kg.s-1), \( g \) is the acceleration due to gravity (m.s-2), and \( h \) is the height the mass is lifted (m).
The output power may now be calculated.
\( P = \dot{m} g h = 5 \cdot 9.81 \cdot 3 = \) = 147.15 W.
The required output power of the pump is 147.15 W. The pump has an efficiency of 80%, i.e. \( \eta = \) 0.8. The standard output power and efficiency equation may be re-arranged to get the input power.
\( P_{\mathrm{out}} = \eta P_{\mathrm{in}} \rightarrow P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} \)
Plugging the known quantities into the equation above yields the answer.
\( P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} = \frac{147.15}{0.8} = \) 183.9 W (4 sf) #.
Question 10
A motor lifts a 10 kg load, 5 m vertically in 3 s. If the motor is 90% efficient calculate the input power.
Solution
In order to calculate the input power, the required output power requires calculation.
The lifting rate is 5 m in 3 s. If the speed is constant, the speed of lifting is 1.66667 ms-1.
The equation for the output power given a constant speed is given below.
\( P = m g v \)
where \( P \) is the rate of work being done (J.s-1 or W), \( m \) is the mass (kg), \( g \) is the acceleration due to gravity (m.s-2), and \( v \) is the speed the mass is lifted (ms-1).
The output power may now be calculated.
\( P = m g v = 10 \cdot 9.81 \cdot 1.66667 = \) = 163.5 W (6 sf).
The required output power of the motor is 163.5 W. The motor has an efficiency of 90%, i.e. \( \eta = \) 0.9. The standard output power and efficiency equation may be re-arranged to get the input power.
\( P_{\mathrm{out}} = \eta P_{\mathrm{in}} \rightarrow P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} \)
Plugging the known quantities into the equation above yields the answer.
\( P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} = \frac{163.5}{0.9} = \) 181.7 W (4 sf) #.
Question 11
A motor lifts a 5.5 kg load up 10 m while rotating at 75 rpm. What is the input power rating of the motor if it is 80% efficient.
Solution
NOTE: This question is not solvable, because the amount if time the mass takes to be lifted is not known. Without this information, it is not possible to calculate the power, and therefore it is not possible to calculate the answer. This "solution" will attempt to reverse-engineer the answer. The answer is given as 5.3 kW.
The energy, or work required to lift a mass parallel to a force is given below.
\( E = m g h \)
where \( E \) is the work done (J), \( m \) is the mass (kg), \( g \) is the acceleration due to gravity (m.s-2), and \( h \) is the height the mass is lifted (m).
The difference between this equation and torque is that torque is always exerted at right angles to the radius. In this sense, a torque does no work. The force in \( E = m g h \) is being exerted parallel to the direction of movement, therefore work is done.
With that known, the work is calculated.
\( E = m g h = 5.5 \cdot 9.81 \cdot 10 = \) 539.55 J (6 sf).
The problem here is that we do not know how long the machine takes to do this amount of work. While you could easily imagine a motor-driven pulley, if the radius is not known, it is not possible to know how many revolutions of the pulley were required to lift the mass 10 m.
Conversely, the radius of the pulley is not known, so it is not possible to know the output torque of the motor. Without knowing the torque, it is impossible to calculate the power output.
Assuming the input power is in fact 5.3 kW, we can attempt to reverse engineer the problem.
The motor takes an input power of 5.3 kW, and has an efficiency of 80%.
\( P_{\mathrm{out}} = \eta P_{\mathrm{in}} = 5300 \cdot 0.8 = \) 4240 W.
The useful output power of the motor is 4240 W.
The power output for a torque is given below.
\( P = \tau \omega \)
where \( P \) is the output power (W), \( \tau \) is the torque (Nm), and \( \omega \) (Greek small letter 'omega') is the angular velocity (rad.s-1). Note that radians are a unit of angle measure. They are similar to degrees except while 360° is one revolution, \( 2 \pi \) radians are one revolution.
However, \( \tau = m g r \), since the force that creates the torque is the weight of the 10 kg mass. The equation with substitutions is shown below.
\( P = m g r \omega \)
This equation can be re-arranged to yield \( r \).
\( P = m g r \omega \rightarrow r = \frac{P}{m g \omega}\)
In order to use this equation, the 75 rpm speed must be converted to rad.s-1.
1 rpm is 1 revolution per minute. Since a "revolution" has no dimension, 1 rpm can be written as 1 "per minute", or 1 min-1.
1 min-1 is also \( \frac{1}{60} \) s-1, which is the same as \( 2 \pi \) rad.s-1, since one revolution is \( 2 \pi \) radians.
Bringing that all together, 1 min-1 is \( \frac{2 \pi}{60} \) rad.s-1.
The rotational speed of 75 rpm therefore becomes \( \omega = 75 \cdot \frac{2 \pi}{60} = \) 7.83598 rad.s-1.
\( r = \frac{4240}{5.5 \cdot 9.81 \cdot 7.83598} = \) 10.0286 m.
In other words, the pulley would have to have had a radius of ~10.03 m for this problem to be solvable.
This also corresponds to an output motor torque of 541.1 Nm, based on the torque the weight of the mass would exert on the pulley wheel.
Question 12
A motor producing 50 Nm of torque at 600 rpm has a 300 mm diameter pulley attached to the output shaft. A drive belt connects this pulley to another pulley having a diameter of 175 mm. Calculate the torque and speed (in rpm) on this smaller pulley.
Solution
The key to this question is to remember that (ideal) pulleys conserve power. They neither product nor consume power. In this sense they are very similar to a transformer in the electrical world.
Two pulleys attached by a belt
will always move the same circumferential distance in the same time. This is also referred to as the no-slip condition.
The size difference of the pulleys will cause them to move at different rotational speeds to preserve the no-slip condition.
The power output for a torque is given below.
\( P = \tau \omega \)
where \( P \) is the output power (W), \( \tau \) is the torque (Nm), and \( \omega \) (Greek small letter 'omega') is the angular velocity (rad.s-1). Note that radians are a unit of angle measure. They are similar to degrees except while 360° is one revolution, \( 2 \pi \) radians are one revolution.
It is usually simpler to work with revolutions per minute than radians per second.
The power equation can be re-written as below.
\( P = \tau \cdot \frac{2 \pi n}{60} \)
where \( P \) is the output power (W), \( \tau \) is the torque (Nm), \( n \) is the rotational speed (rpm).
This equation can be used to relate the speeds and torques of two pulleys attached by a belt.
\( P_1 = P_2 \rightarrow \tau_1 \cdot \frac{2 \pi n_1}{60} = \tau_2 \cdot \frac{2 \pi n_2}{60} \rightarrow \tau_1 n_1 = \tau_2 n_2 \)
The no-slip condition is also useful. The circumferential speed is given below.
\( v = r \omega \)
where \( v \) is the circumferential velocity (ms-1), and \( \omega \) (Greek small letter 'omega') is the angular velocity (rad.s-1).
This equation can be used to relate the radii and speeds of two pulleys attached by a belt.
\( v_1 = v_2 \rightarrow r_1 \cdot \frac{2 \pi n_1}{60} = r_2 \cdot \frac{2 \pi n_2}{60} \rightarrow r_1 n_1 = r_2 n_2 \)
This equation also works for diameters, since diameter is simply twice the radius.
\( r_1 n_1 = r_2 n_2 \rightarrow d_1 n_1 = d_2 n_2 \)
With that known, the problem can be solved.
The first pulley has \( d_1 \) = 300 mm, and \( n_1 \) = 600 rpm. The second pulley has an unknown speed but a diameter of 175 mm. The speed is able to be calculated.
\( d_1 n_1 = d_2 n_2 \rightarrow 300 \cdot 600 = 175 \cdot n_2 \rightarrow n_2 = \frac{300 \cdot 600}{175} = \frac{180000}{175} = \) 1028.57 (6 sf) or 1029 rpm (4 sf) #.
The torque is also calculated.
\( \tau_1 n_1 = \tau_2 n_2 \rightarrow 50 \cdot 600 = \tau_2 \cdot 1028.57 \rightarrow \tau_2 = \frac{50 \cdot 600}{1028.57} = \frac{30000}{1028.57} = \) 29.17 Nm (4 sf) #.
Question 13
Sketch a Hypoid bevel-tooth gear.
Solution
I don't know how much detail is expected of the answer. The hypoid gear has an axis of rotation for the pinion gear (small gear) that does not intercept the axis of rotation of the ring gear. See below.
Question 14
When are drive gears used instead of drive belts.
Solution
Drive gears are used where the power requirements make belts difficult to use and/or inefficient.
It may also depend on packaging. A gearbox is better when the output is to be inline with the input.
The actual choice is application driven.
Question 15
A 5kW motor (input) having a 90% efficiency rating, drives a generator of 60% efficiency. Calculate the maximum generator output.
Solution
This question is an example of chained efficiency, where energy goes through a chain of transformations from an initial form to a final form.
The efficiency of a chain of transformations is the product of all the steps in that chain.
\( \eta_{\mathrm{TOT}} = \eta_1 \eta_2 \eta_3 \ldots \)
where \( \eta_{\mathrm{TOT}} \) is the efficiency of the entire chain, and \( \eta_n \) is the efficiency of step \( n \). Note that this formula only works if all the efficiencies are handled as numbers from 0 to 1.
For this situation, \( P_{\mathrm{in}} \) is 5 kW, and \( \eta_{\mathrm{TOT}} = 0.9 \cdot 0.6 = \) 0.54.
Calculating the output power is now straightforward.
\( P_{\mathrm{out}} = \eta P_{\mathrm{in}} = 0.54 \cdot 5 = \) 2.7 kW #.
Question 16
The reading on a kWh meter over 2 hours is 13.5 units. Calculate the average power. How many joules have been used in total?
Solution
1 unit is 1 kWh.
The meter has read 13.5 kWh in 2 h. The average power is calculated.
\( P = \frac{E}{t} = \frac{13.5}{2} = \) 6.75 kW #.
Note is was not necessary to convert any units. A kWh (kilowatt-hour) quantity divided by an h (hour) quantity will yield a kW (kilowatt) quantity.
Converting to joules is easy, remembering that 1 kWh = 3.6 MJ.
\( E = 13.5 \cdot 3600000 = \) 48600000 J or 46.8 MJ #.
Question 17
Calculate the force applied if a 2kg mass is accelerated at a rate of 10 ms-2.
Solution
This question uses \( F = ma \).
\( F = m a = 2 \cdot 10 = \) 20 N #.
Question 18
A 5kW motor runs for 5 hours. Calculate total kWhs, total Watt-seconds, and total joules.
Solution
For all calculations, the equation \( E = P t \) is used.
\( E = P t = 5 \cdot 5 = \) 25 kWh #.
\( E = 25 \cdot 3.6 = \) 90 MJ #.
1 watt-second is 1 joule.
\( E = \) 90 MWs #.
Question 19
A 250kg load is to be raised at a speed of 5 ms-1. The gearing is 90% efficient and the motor is 70% efficient. Calculate the input power to the motor.
Solution
First, the required output power is calculated.
\( P_{\mathrm{out}} = m g v = 250 \cdot 9.81 \cdot 5 = \) 12262.5 W (6 sf).
The efficiency of the entire system is the product of the motor and gearing efficiencies. This yields an \( \eta \) of 63% or 0.63.
The input power is now straightforward to calculate.
\( P_{\mathrm{in}} = \frac{P_{\mathrm{out}}}{\eta} = \frac{12262.5}{0.63} = \) 19.46 kW (4 sf) #.
Question 20
A pump delivers 250 litres of water from a 20m deep trench every 5 seconds. The pump runs for a total of 5 hours.
Calculate:
- The total energy used in joules, watt-seconds and kWh.
- Minimum power rating of the motor.
- Total cost to pump out all the water.
NOTE: You do not have to do the calculations in the same order as the results!
Solution
The equation for the output power given a constant rate of mass transfer is given below.
\( P = \dot{m} g h \)
where \( P \) is the rate of work being done (J.s-1 or W), \( \dot{m} \) is the mass transfer rate (kg.s-1), \( g \) is the acceleration due to gravity (m.s-2), and \( h \) is the height the mass is lifted (m).
The pump transfers 250 L of water every 5 s. This translates to a volume flow rate of 50 L.s-1. Given that water has a density of 1 kg.L-1, the mass transfer rate is 50 kg.s-1.
The height is 20 m. This is enough information to calculate the power output of the motor.
\( P = \dot{m} g h = 50 \cdot 9.81 \cdot 20 = \) 9810 W or 9.81 kW.
The motor runs for five hours. The energy use is calcualted.
\( E = P t = 9.81 \cdot 5 = \) 49.05 kWh.
\( E = 49.05 \cdot 3.6 = \) 176.58 MJ #.
1 watt-second is 1 joule.
\( E = \) 176.58 MWs #.
a) Total Energy etc.
The total energy is:
- 176.58 MJ
- 176.58 MWs
- 49.05 kWh
b) Minimum Power Rating
The minimum motor power rating is 9.81 kW.
Note that this figure does not take into account motor losses!
c) Cost of Running
The electricity used to run the motor has a $0.25 tariff. The cost is calculated below.
\( C = r E = 0.25 \cdot 49.05 = \) $12.26 #.
Note that this figure does not take into account motor losses!