Category Archives: Engineering Mathematics Blog

Hibbeler Statics 14E P1.4 — Conversion from English units to Metric Units

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

_{Statics of Rigid Bodies 14^th by RC Hibbeler, Problem 1-4}

Solution:

Part A

\begin{align*} 200\ \text{lb}\cdot \text{ft} & =\left( 200\ \text{lb}\cdot \text{ft} \right)\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\\ &=271\ \text{N}\cdot \text{m} \end{align*}

Part B

\begin{align*} 350 \ \text{lb/ft}^3& = \left( \frac{350\ \text{lb}}{1\ \text{ft}^3} \right)\left( \frac{1\ \text{ft}}{0.3048\ \text{m}} \right)^3\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{1\ \text{kN}}{1000\ \text{N}} \right)\\ & = 55.0\ \text{kN/m}^3 \end{align*}

Part C

\begin{align*} 8 \ \text{ft/hr}& = \left( \frac{8\ \text{ft}}{1\ \text{hr}} \right)\left( \frac{1\ \text{hr}}{3600\ \text{s}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\left( \frac{1000 \ \text{mm}}{1 \ \text{m}} \right)\\ & = 0.677\ \text{mm/s} \end{align*}

Hibbeler Statics 14E P1.3 — Representing a combination of units in the correct SI form

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/ms, (b) N/mm, and (c) mN/(kg・µs).

_{Statics of Rigid Bodies 14^th Edition by RC Hibbeler, Problem 1-3}

Solution:

Part A

\begin{align*} \text{Mg/ms} & = \frac{10^3 \ \text{kg}}{10^{-3} \ \text{s}} \\ & = 10^6 \text{kg/s}\\ & = \text{Gg/s} \end{align*}

Part B

\begin{align*} \text{N/mm} & = \frac{1\ \text{N}}{10^{-3} \ \text{m}}\\ & = 10^3 \ \text{N/m}\\ & = \text{kN/m} \end{align*}

Part C

\begin{align*} \frac{\text{mN}}{\left( \text{kg} \cdot \mu \text{s} \right)} & = \frac{10^{-3} \ \text{N}}{10^{-6} \ \text{kg} \cdot \text{s}}\\ & =10^3 \ \text{N}/\left( \text{kg} \cdot \text{s} \right)\\ & = \text{kN}/\left( \text{kg} \cdot \text{s}\right) \end{align*}

Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is $\mu _s=0.60$ ; (b) the pavement is icy and $\mu _s=0.25$ .

Solution:

The forces on the car are gravity mg downward, the normal force F_N exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

\begin{align*} \sum_{}^{}F_y & =m\ a_c\\ \\ F_N\ -\ mg & =0\\ \\ F_N&=mg\\ \\ F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\ \\ F_N &=9810\ \text{N} \end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

\begin{align*} \sum_{}^{}F_c & =m\ a_c\\ \\ & =m\cdot \frac{v^2}{r}\\ \\ & =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\ \\ &=4500\ \text{N} \end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), $\mu _s=0.60$ , and the maximum friction force attainable is

\begin{align*} \sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\ \\ &=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\ \\ &=5886\ \text{N} \end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

\begin{align*} \sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\ \\ &=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\ \\ &=2452.5\ \text{N} \end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.

Revolving Ball (Vertical Circle) – Uniform Circular Motion Example Problem

A 0.150-kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. (b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (a).

Solution:

The ball moves in a vertical circle and is not undergoing uniform circular motion. The radius is assumed constant, but the speed v changes because of gravity. Nonetheless, the equation for centripetal acceleration $\text{a}_\text{c} = \frac{\text{v}^2}{\text{r}}$ is valid at each point along the circle, and we use it at the top and bottom points. The free body diagram is shown in the figure below for both positions.

Part A

At the top (point 1), two forces act on the ball: mg, the force of gravity (or weight), and F_T1, the tension force the cord exerts at point 1. Both act downward, and their vector sum acts to give the ball its centripetal acceleration a_c. We apply Newton’s second law, for the vertical direction, choosing downward as positive since the acceleration is downward (toward the center):

\begin{align*} \sum_{}^{}\text{F}_\text{v}& =\text{ma}_\text{c}\\ \\ \text{F}_\text{T1}\ +\ \text{mg}&= \text{m} \cdot \frac{\text{v}_1^2}{\text{r}} \end{align*}

From this equation we can see that the tension force F_T1at point 1 will get larger if v₁ (ball’s speed at top of circle) is made larger, as expected. But we are asked for the minimum speed to keep the ball moving in a circle. The cord will remain taut as long as there is tension in it. But if the tension disappears (because v₁ is too small) the cord can go limp, and the ball will fall out of its circular path. Thus, the minimum speed will occur if F_T1 = 0 (the ball at the topmost point), for which the equation above becomes

\text{mg}=\text{m}\cdot \frac{\left( \text{v}_1 \right)^2}{\text{r}}

We solve for v₁, we have

\begin{align*} \text{v}_1&=\sqrt{\text{gr}} \\ \\ &=\sqrt{\left( 9.81\ \text{m/s}^2 \right)\left( 1.10\ \text{m} \right)} \\ \\ &=3.285 \ \text{m/s} \end{align*}

Therefore, the minimum speed at the top of the circle if the ball is to continue moving in a circular path is about 3.285 m/s.

Part B

When the ball is at the bottom of the circle, the cord exerts its tension force F_T2 upward, whereas the force of gravity, mg still acts downward. Choosing upward as positive, Newton’s second law gives:

\begin{align*} \sum_{}^{}\text{F}_\text{v}& =\text{ma}_\text{c}\\ \\ \text{F}_\text{T2}\ -\ \text{mg}&= \text{m} \cdot \frac{\text{v}_2 ^2}{\text{r}} \end{align*}

The speed v₂ is given as twice that in (a). We solve for F_T2

\begin{align*} F_\text{T2} & = m\cdot \frac{v^2}{r}+mg\\ \\ & = \left( 0.150\ \text{kg} \right)\cdot \frac{\left( 2\times 3.285 \ \text{m/s}\right)^2}{1.10\ \text{m}}+\left( 0.150\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\ \\ &=7.358 \ \text{N} \end{align*}

Force on Revolving Ball (Horizontal) – Uniform Circular Motion Example Problem

Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

Solution:

First we need to draw the free-body diagram for the ball. The forces acting on the ball are the force of gravity (or weight), mg downward, and the tension force F_T that the string exerts toward the hand at the center (which occurs because the person exerts that same force on the string). The free-body diagram for the ball is shown in the figure below. The ball’s weight complicates matters and makes it impossible to revolve a ball with the cord perfectly horizontal. We estimate the force assuming the weight is small, and letting $\phi = 0$ from the figure. Then F_T will act nearly horizontally and, in any case, provides the force necessary to give the ball its centripetal acceleration.

Before, we can use the formula of the centripetal force, we need to solve for the value of the linear velocity first. The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*} \text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\ \\ & = 7.54 \ \text{m/s} \end{align*}

Using the formula for centripetal force, we have

\begin{align*} \text{F}_\text{c} &=\text{ma}_\text{c} \\ \\ & = \text{m} \cdot \frac{\text{v}^{2}}{\text{r}} \\ \\ & = \left( 0.150\ \text{kg} \right) \cdot \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\ \\ & = 14.2\ \text{N} \end{align*}

Therefore, the force a person must exert on a string is about 14.2 N.

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*} \text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi \text{r}}{\text{t}} \\ \\ & = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\ \\ & = 7.54 \ \text{m/s} \end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, a_c.

\begin{align*} \text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\ \\ & = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\ \\ & =94.8 \ \text{m/s}^{2} \end{align*}

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Physics Problem: Fast-Plant

Bamboo grows very fast, for a plant. A particular bamboo might grow 90.4 cm in a single day. How many micrometers per second does this rate correspond to?

Solution:

\begin{aligned} \text{rate} & = \frac{\text{length}}{\text{time}} \\ \\ &=\frac{90.4\ \text{cm}}{1 \ \text{day}}\\ \\ &=\frac{90.4\ \bcancel{\text{cm}}}{\bcancel{\text{day}}} \times\frac{10,000\ \mu m}{1\ \bcancel{\text{cm}}}\times \frac{1\ \bcancel{\text{day}}}{86400\ s}\\ \\ & = 10.5\ \mu m/s \end{aligned}

College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, $y_0$ of the rock is 105 meters, and the initial velocity, $v_0$ is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned} \Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\ &=\left( 0 \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\ &=0+11.036\ \text{m} \\ &=11.04 \ \text{m} \end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned} \text{y}&=\text{y}_{0}-\Delta \text{y} \\ &=105\ \text{m}-11.04\ \text{m} \\ &=93.96 \ \text{m} \end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned} \text{y} & =\frac{1}{2}\text{at}^{2} \\ &\text {Solving for t, we have}\\ \text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\ &=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\ &=4.63 \ \text{s} \end{aligned}

Therefore, to move out the hiker has about

\begin{aligned} \text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\ &=3.13\ \text{s} \end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\ \end{aligned}

we have

\begin{aligned} \text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\ \text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\ \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \end{aligned}

Substituting the known values,

\begin{aligned} \text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\ \text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\ \text{v}_0&= {\color{green}7.00 \ \text{m/s}} \end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so $y$ is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s².

\begin{aligned} \text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\ 0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\ 0 & =7\text{t}-4.905\text{t}^{2}\\ 7\text{t}-4.905\text{t}^{2}&=0 \\ \text{t}\left( 7-4.905\text{t} \right) & =0 \\ \text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\ \end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned} 7-4.905\text{t} &=0 \\ 4.905\text{t} & = 7 \\ \text{t} & =\frac{7}{4.905} \\ \text{t}&={\color{green}1.43 \ \text{s}} \end{aligned}

The kangaroo is about 1.43 seconds long in the air.

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Hibbeler Statics 14E P1.4 — Conversion from English units to Metric Units

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

Hibbeler Statics 14E P1.3 — Representing a combination of units in the correct SI form

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/ms, (b) N/mm, and (c) mN/(kg・µs).

Part C

Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is $\mu _s=0.60$ ; (b) the pavement is icy and $\mu _s=0.25$ .

Revolving Ball (Vertical Circle) – Uniform Circular Motion Example Problem

Force on Revolving Ball (Horizontal) – Uniform Circular Motion Example Problem

Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Purchase Textbook: Differential and Integral Calculus by Feliciano and Uy

Physics Problem: Fast-Plant

Bamboo grows very fast, for a plant. A particular bamboo might grow 90.4 cm in a single day. How many micrometers per second does this rate correspond to?

College Physics 2.51 – Time of the hiker to move out from a falling rock

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft3 to kN/m3, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/ms, (b) N/mm, and (c) mN/(kg・µs).

Part C

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is μs=0.60\mu _s=0.60μs​=0.60; (b) the pavement is icy and μs=0.25\mu _s=0.25μs​=0.25.

Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. Ignore the string’s mass.

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Bamboo grows very fast, for a plant. A particular bamboo might grow 90.4 cm in a single day. How many micrometers per second does this rate correspond to?

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is $\mu _s=0.60$ ; (b) the pavement is icy and $\mu _s=0.25$ .