202205311653 Exercise 1W

When a=1, b=2, c=5, d=6, and m=0, give the value of

1. ad^2+5bc 2. a^2bc^2-3ad+a^4 3. mabc+3a^2b^2c^2 4. b^2cd-abc^2 5. a^4b^3c+bcd 6. ma^3+4cd-ad 7. \displaystyle{\frac{ab^2}{2}+5c} 8. \displaystyle{\frac{md}{3}+\frac{4cd}{b}} 9. \displaystyle{\frac{ac}{d}+a} 10. \displaystyle{\frac{a^2}{c}+3a^2b} 11. \displaystyle{\frac{ab^2}{a}-\frac{a^5d}{b}} 12. \displaystyle{\frac{c+d}{d}+\frac{d-c}{a}} 13. \displaystyle{\frac{d^2-a^2b^2}{a^2b^2}} 14. \displaystyle{\frac{mcd+3abc}{d^2-4b^2}} 15. \displaystyle{\frac{a^2d^2-9b^2}{abcd}}

J. R. Lux & R. S. Pieters. (1969). Exercises in Elementary Algebra


Solution.

1.

\begin{aligned} ad^2+5bc & = (1)(6)^2+5(2)(5) \\ & = 36 + 50 \\ & = 86 \\ \end{aligned}

2.

\begin{aligned} a^2bc^2-3ad+a^4 & = (1)^2(2)(5)^2-3(1)(6)+(1)^4 \\ & = 50-18+1 \\ & = 33 \\ \end{aligned}

3.

\begin{aligned} mabc+3a^2b^2c^2 & = (0)(1)(2)(5)+3(1)^2(2)^2(5)^2 \\ & = 0 + 3(4)(25) \\ & = 300 \\ \end{aligned}

4.

\begin{aligned} b^2cd-abc^2 & = (2)^2(5)(6)-(1)(2)(5)^2 \\ & = (4)(5)(6)-(2)(25) \\ & = 120-50 \\ & = 70 \\ \end{aligned}

5.

\begin{aligned} a^4b^3c+bcd & = (1)^4(2)^3(5)+(2)(5)(6) \\ & = 8(5) + 60 \\ & = 40 + 60 \\ & = 100 \\ \end{aligned}

6.

\begin{aligned} ma^3+4cd-ad & = (0)(1)^3+4(5)(6)-(1)(6) \\ & = 0 + 120 - 6 \\ & = 114 \\ \end{aligned}

7.

\begin{aligned} \frac{ab^2}{2} + 5c & = \frac{(1)(2)^2}{2}+5(5) \\ & = \frac{4}{2} + 25 \\ & = 2 + 25 \\ & = 27 \\ \end{aligned}

8.

\begin{aligned} \frac{md}{3}+\frac{4cd}{b} & = \frac{(0)(6)}{3}+\frac{4(5)(6)}{(2)} \\ & = 0 + \frac{120}{2} \\ & = 60 \\ \end{aligned}

9.

\begin{aligned} \frac{ac}{d}+a & = \frac{(1)(5)}{(6)} + (1) \\ & = \frac{5}{6} + 1 \\ & = 1\frac{5}{6} \\ \end{aligned}

10.

\begin{aligned} \frac{a^2}{c}+3a^2b & = \frac{(1)^2}{(5)}+3(1)^2(2) \\ & = \frac{1}{5}+6 \\ & = 6\frac{1}{5} \\ \end{aligned}

11.

\begin{aligned} \frac{ab^2}{a}-\frac{a^5d}{b} & = \frac{(1)(2)^2}{(1)}-\frac{(1)^5(6)}{(2)} \\ & = 4 - 3 \\ & = 1 \\ \end{aligned}

12.

\begin{aligned} \frac{c+d}{d}+\frac{d-c}{a} & = \frac{(5)+(6)}{(6)} + \frac{(6)-(5)}{(1)} \\ & = \frac{11}{6} + \frac{1}{1} \\ & = 1\frac{5}{6} + 1 \\ & = 2\frac{5}{6} \\ \end{aligned}

13.

\begin{aligned} \frac{d^2-a^2b^2}{a^2b^2} & = \frac{(6)^2-(1)^2(2)^2}{(1)^2(2)^2} \\ & = \frac{36-(1)(4)}{(1)(4)} \\ & = \frac{32}{4} \\ & = 8 \\ \end{aligned}

14.

\begin{aligned} \frac{mcd+3abc}{d^2-4b^2} & = \frac{(0)(5)(6)+3(1)(2)(5)}{(6)^2-4(2)^2} \\ & = \frac{0+30}{36-4(4)} \\ & = \frac{30}{36-16} \\ & = \frac{30}{20} \\ & = \frac{3}{2} \\ & = 1\frac{1}{2} \\ \end{aligned}

15.

\begin{aligned} \frac{a^2d^2-9b^2}{abcd} & = \frac{(1)^2(6)^2-9(2)^2}{(1)(2)(5)(6)} \\ & = \frac{(1)(36)-9(4)}{60} \\ & = \frac{36-36}{60} \\ & = \frac{0}{60} \\ & = 0 \\ \end{aligned}

202205311107 Problem 4.23

Boxes A and B are in contact on a horizontal, frictionless (i.e. f=0) surface, as shown in the Figure below. Box A has mass 20.0\,\mathrm{kg} and box B has mass 5.0\,\mathrm{kg}. A horizontal force of 100\,\mathrm{N} is exerted on box A. What is the magnitude of the force that box A exerts on box B?

extracted from Problem 4.23, Sears and Zemansky’s University Physics


Steps.

Draw the free-body diagrams of A, B, and A+B.

Apply Newton’s 2^\textrm{nd} law \textrm{Net }F=ma:

\begin{aligned} F_A- {}_{B}F_A - f_A & = m_Aa_A \\ {}_{A}F_B - f_B & = m_Ba_B \\ F_A-f_{A+B} & = m_{A+B}a_{A+B} \\ \end{aligned}

Conditioning the equations of motion:

\begin{aligned} a_A=a_B & =a_{A+B} \\ f_A=f_B=f_{A+B} & =0 \\ {}_{B}F_A & ={}_{A}F_{B} \\ m_A+m_B & =m_{A+B} \\ \end{aligned}

and substituting numbers for symbols, write:

\begin{aligned} 100 - {}_{A}F_{B} & = 20a \\ {}_{A}F_{B} & = 5a \\ 100 & =25a \\ \end{aligned}


\therefore The magnitude {}_{A}F_{B} of the force that box A exerts on box B is 20\,\mathrm{N}.