Solutions to Tutorial Lesson 1. mechanical engineering principle-tutorial. covering basic questions...

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SOLUTIONS TO TUTORIAL SELECTION LECTURE 1

Multiple choice:

Q2

Suppose two quantities, A and B, have different dimensions. Determine which of the following arithmetic operations could be physically meaningful. (a) A + B (b) B - A (c) A - B (d) A/B (e) AB

Since A and B have different dimensions, they cannot be simply added or subtracted...hence (a), (b), and c are not physically meaningful. However (d) and (e) could be.

Q3

Newton’s second law of motion (Chapter 4) says that the mass of an object times its acceleration is equal to the net force on the object. Which of the following gives the correct units for force? (a) kg m/s2 (b) kg m2/s2 (c) kg/m2 s2 (d) kg m2/s (e) none of these.

We know that Force = mass x acceleration, hence part (a) must be correct.

Q6

A house is advertised as having 1 420 square feet under roof. What is the area of this house in square metres? (a) 115 m2 (b) 132 m2 (c) 176 m2 (d) 222 m2 (e) none of these.

We need to convert ft to m. We know that 1 m = 3.281 ft (Serway frontplate), so

22 2

2 21 m1420 ft 131.91m

3.281 ft

⎡ ⎤⎡ ⎤ × =⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

So 132 m2 is the closest answer

We could make an estimate…3.2812 is close to 10 (actually 10.76) hence area ≅ 142 m2, so 132 is the closest

Q9

Which of the following relationships is dimensionally consistent with an expression yielding a value for acceleration? Acceleration has the units of distance divided by time squared. In these equations, x is a distance, t is time, and v is velocity with units of distance divided by time. (a) v/t2 (b) v/x2 (c) v2/t (d) v2/x (e) none of these.

We need to do a dimensional analysis on each one and see if it has the dimensions L/T2

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2

2 3v L 1 L(a) . .....so not acceleration!

T Tt T⎡ ⎤→ →⎢ ⎥⎣ ⎦

2

2v L 1 1(b) . .....so not acceleration!

T L LTx⎡ ⎤→ →⎢ ⎥⎣ ⎦

22 2

3v L 1 L(c) . .....so not acceleration!t T T T

⎡ ⎤ ⎡ ⎤→ →⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

22

2v L 1 L(d) . .....sodimensionssuggest acceleration!x T L T

⎡ ⎤ ⎡ ⎤→ →⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Q10

The price of gasoline at a particular station is 1.5 euros per litre. An American student can use 33 euros to buy gasoline. Knowing that 4 US quarts make a US gallon and that one litre is close to one US quart, she quickly reasons that she can buy how many US gallons of gasoline? (a) less than 1 US gallon (b) about 5 US gallons (c) about 8 US gallons (d) more than 10 US gallons

Note that 4 quarts = 1 gallon is also true for British units…they are just a different size (bigger) than US values

[ ] 33 euro / litreShe can buy = 22 litres1.5 euro

We now need to convert this to US gallons

[ ] USquart 1 USgallon22 litre 1 5.5 USgallonslitre 4 USquart

⎡ ⎤⎡ ⎤× × =⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

So answer (b) is the closest

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Q11

At a horizontal distance of 45 m from a tree, the angle of elevation to the top of the tree is 26°. How tall is the tree? (a) 22 m (b) 31 m (c) 45 m (d) 16 m (e) 11 m

Probably it’s best to make a sketch

To solve this we need to use trig

opposite side hTan 26adjacent side 45

= =

then h 45 tan 26 45 0.4877 21.95m= × = × =

The closest answer is (a) 22 m

Problems:

Q1

The period of a simple pendulum, defined as the time necessary for one complete

oscillation, is measured in time units and is given by t 2g

= π

where ℓ is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent.

To solve this carry out a dimensional analysis…we know the answer should be T

Also 2π is dimensionless.

22T LL T L hence it is dimensionally consistent

g L L⎡ ⎤ /⎡ ⎤→ → →⎢ ⎥ ⎢ ⎥/⎣ ⎦⎢ ⎥⎣ ⎦

45 m

260

Height h

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Q2 (a)Suppose that the displacement of an object is related to time according to the expression x = Bt2. What are the dimensions of B?

(b) A displacement is related to time as x = A sin (2πft), where A and f are constants. Find the dimensions of A. (Hint: A trigonometric function appearing in an equation must be dimensionless.)

(a) First write B in terms of x and t2

2 2x LBt T

⎡ ⎤= → ⎢ ⎥⎣ ⎦

(b) Follow a similar approach noting that sin (2πft) is dimensionless. Thus x must have the same dimensions as A, i.e. L

Q3

A shape that covers an area A and has a uniform height h has a volume V = Ah.

(a) Show that V = Ah is dimensionally correct.

(b) Show that the volumes of a cylinder and of a rectangular box can be written in the form V = Ah, identifying A in each case.

(Note that A, sometimes called the “footprint” of the object, can have any shape and that the height can, in general, be replaced by the average thickness of the object.)

(a) The dimensions of V are L3 and those of A x h = L2 x L →L3 ie V = Ah is dimensionally correct.

h

A A

h

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Q4

Each of the following equations was given by a student during an examination:

(a) mghmv21

mv21 2

02 +=

(b) v = v0 + at2

(c) ma = v2.

Do a dimensional analysis of each equation and explain why the equation can’t be correct.

2 20

1 1(a) mv mv mgh2 2

= +

We can see that the dimensions of 2 2

2 20 2

1 1 L MLmv and mv are the sameandare : M2 2 T T

⎡ ⎤⎡ ⎤ →⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

2 12

2L L LThe dimensions of mgh are M L M M

T TT⎡ ⎤→ →⎢ ⎥⎣ ⎦

Hence the equation must be incorrect

(b) v = v0 + at2

We need to see if the dimensions of at2 are the same as those of v

2 22

L LWe know dimensions of v are then at T LT T

→ →

Again the equation must be incorrect

(c) ma = v2

22

2 2L LDimensions of ma are M and those of v areT T

Yet again the equation must be incorrect

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Q5

Newton’s law of universal gravitation is represented by 2rMm

GF =

where F is the gravitational force, M and m are masses, and r is a length. Force has the SI units kg m/s2. What are the SI units of the proportionality constant G?

First rearrange the equation to make G the subject of the equation

2

2Mm FrF G then G

Mmr= =

Now put in the SI base units

2 2 2 3

2 2 2 2Fr Nm kgm m mGMm kg s kg kgs

= → → →

Q19

A car is travelling at a speed of 38.0 m/s on an inter- state highway where the speed limit is 75.0 mile/h. Is the driver exceeding the speed limit? Justify your answer. All we have to do here is convert metres to miles and seconds to hours. We know 1 mile = 1.609 km = 1609 m and 1 hour = 3600 seconds

m m mile 3600s 38 3600 mile milev 38 38 85.021s s 1609m h 1609 h h

⎡ ⎤ ×⎡ ⎤ ⎡ / ⎤ ⎡ ⎤ ⎡ ⎤/= → × × → =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥//⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

So s/he is exceeding the speed limit by 10 mile/h Q20

A certain car has a fuel efficiency of 25.0 miles per gallon (mi/gal). Express this efficiency in kilometres per litre (km/L)

Note that the answer will be different if we use UK gallons or US gallons. I’ve chosen UK gallons where 1 gallon = 4.54609 litres

mile 1.609 km gallon 1.609 km 25 1.609 km km25 25 8.848gallon mile 4.5461litre 4.5461 litre 4.5461 litre litre⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ×⎡ ⎤ ⎡ ⎤× × → = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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If I had chosen US gallons where 3.786 litre = 1 US gallon we would get

mile 1.609 km gallon 1.609 km 25 1.609 km km25 25 10.625gallon mile 4.5461litre 3.786 litre 3.786 litre litre⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ×⎡ ⎤ ⎡ ⎤× × → = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Q21

The diameter of a sphere is measured to be 5.36 in. Find (a) the radius of the sphere in centimetres, (b) the surface area of the sphere in square centimetres, and (c) the volume of the sphere in cubic centimetres.

First calculate the diameter in cm where 1 inch = 2.54 cm

[ ] cm5.36 inch 2.54 13.614cminch

⎡ ⎤× =⎢ ⎥⎣ ⎦

It is convenient to use the radius = half the diameter = 6.807 cm

Now we need the equations for surface area and volume

2 2 2 2 2 2Surface area A 4 r 4 6.807 cm 582.3 cm or 5.823 10 cm= = π = × π× = ×

3 3 3 3 3 34 4Volume V r 6.807 cm 1321.28 cm or 1.32128 10 cm3 3

= = π = ×π× = ×

Q25

The amount of water in reservoirs is often measured in acre-ft. One acre-ft is a volume that covers an area of one acre to a depth of one foot. An acre is 43 560 ft2. Find the volume in SI units of a reservoir containing 25.0 acre-ft of water.

This question is about converting units. We are told 1 acre is 43 560 ft2 and we know that 1 m = 3.281ft, so 1 acre is….

[ ] [ ]2 2

22 2

ft 1 m1 acre 1 acre 43560 4046.46macre 3.281 ft

⎡ ⎤ ⎡ ⎤→ × × =⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Now convert acre ft…

[ ] [ ]2

2 3m 1 m25 acre ft 25 acre ft 4046.46m 30832.53macre 3.281 ft⎡ ⎤ ⎡ ⎤→ × × =⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

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Q35

A point is located in a polar coordinate system by the coordinates r = 2.5 m and θ = 35°. Find the x- and y-coordinates of this point, assuming that the two coordinate systems have the same origin.

It’s probably best to draw a graph…

Now we can use trig to find y1 and x1

11

ysin 35 hence y 2.5sin35 1.434m2.5

= = =

11

xcos35 hence x 2.5cos35 2.048m2.5

= = =

y axis

x axis

r = 2.5 m

θ = 350

y1

x1

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Q38

Two points in a rectangular coordinate system have the coordinates (5.0, 3.0) and (- 3.0, 4.0), where the units are centimetres. Determine the distance between these points. Again, it is best to draw a graph to visualise what the problem is…

We can see that we have a right angled triangle with the horizontal distance (5 + 3) cm and vertical distance (4 – 3) cm Now use Pythagoras

2 2a 8 1 65 8.062cm= + = =

y axis

x axis

5 cm

x1

• 5.0, 3.0

-3.0, 4.0 •

3 cm 4 cm

3 cm

a cm

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Q39 Two points are given in polar coordinates by (r1 θ1) = (2.00 m, 50.0°) and (r2 θ2) = (5.00 m, -50.0°), respectively. What is the distance between them? Once again draw a graph

Once again we have a right angled triangle…..vertical side = y1 + y2…horz side = x2 – x1

To solve it we need to find these values (neglecting the signs of y) y1 = 2 sin 50 = 1.532 m y2 = 5 sin 50 = 3.83 m hence y1 + y2 = 5.462 m x1 = 2 cos 50 = 1.286 m x2 = 5 cos 50 = 3.214 m hence x2 – x1 = 1.928 m Now use Pythagoras to find a

2 2a 5.462 1.928 5.698m= + =

2 m

5 m

y1

y2

x1

x2

500

-500

a

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Q41

For the triangle shown, what are (a) the length of the unknown side, (b) the tangent of θ, and (c) the sine of φ?

To find the length of the unknown side (x) use Pythagoras:

2 2x 9 6 6.708m= − =

Now just use the definitions of tangent and sine

opposite side 6tan 0.8944adjacent side 6.708

θ = = =

opposite side 6.708sin 0.7453hypo tanuse 9

φ = = =

Q42

A ladder 9.00 m long leans against the side of a building. If the ladder is inclined at an angle of 75.0° to the horizontal, what is the horizontal distance from the bottom of the ladder to the building?

adjacent sideThen use cos hypotenuse

θ = then x = 9 cos 75 = 2.329 m

9 m

750

x m

θ

6 m

9 m φ

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Q43

A high fountain of water is located at the centre of a circular pool. Not wishing to get his feet wet, a student walks around the pool and measures its circumference to be 15.0 m. Next, the student stands at the edge of the pool and uses a protractor to gauge the angle of elevation at the bottom of the fountain to be 55.0°. How high is the fountain?

We know that the circumference of a circle = 2 π r, hence r = 15/2 π = 2.387 m

hNow use tan55= so h r tan55 2.387 tan55 3.409mr

= = × =

Q46

In a certain right triangle, the two sides that are perpendicular to each other are 5.00 m and 7.00 m long. What is the length of the third side of the triangle?

2 2Again apply pythagoras a 5 7 8.602m= + =

5 m

7 m a m

θ

h

r

550

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Q47

In Problem 46, what is the tangent of the angle for which 5.00 m is the opposite side?

tan θ = 5/7 = 0.714

Q55

The displacement of an object moving under uniform acceleration is some function of time and the acceleration. Suppose we write this displacement as nmtkas = where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if m = 1 and n = 2. Can the analysis give the value of k? If these constants are correct, the equation must be dimensionally correct Hence the dimensions of the left hand side [L] must be the same as that of the right hand side.

[ ]m n 1 2 22

La t a t T LT⎡ ⎤ ⎡ ⎤= → →⎢ ⎥ ⎣ ⎦⎣ ⎦

This is dimensionally correct. Note that we assumed k was a dimensionless constant. This analysis does not give us a value for k; to find that we would need further work…probably experiments! Q59

Assume that there are 100 million passenger cars in the United States and that the average fuel consumption is 20 mile/USgal of gasoline. If the average distance travelled by each car is 10 000 mile/yr, how much gasoline would be saved per year if average fuel consumption could be increased to 25 mile/USgal?

Calculate the fuel used each year for each case…

[ ]6 3 10mile 1 galfuelused 100 10 cars 10 10 5 10 galcar 20 mile

⎡ ⎤= × × × × = ×⎢ ⎥⎣ ⎦

Now we can simply ratio this answer by (20/25) so new value = 4 x 1010 gal

Thus savings are 1 x 1010 gallons per year

Solutions to Tutorial Lesson 1. mechanical engineering principle-tutorial. covering basic questions...

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