Applied-math-problem-set.docx

ENGINEERING MECHANICS Problem 1. Find the value of P and F so that the four forces shown in the figure produce an upward resultant of 1335 N acting at 1.20 m from the left end of the bar.

Solution: Take ƩFy = R -440 N+ P – F + 880 N = 1335 N P – F = 1335 + 440 -880 P – F = 895 N

Take moment at F, -440(1.5) + P(0.9) – 880(0.6) = 1335(0.3) 1335(0.3)+ 440(1.5)+ 880(0.6) 0.9

Take moment at P,

P=

440(0.6) – F(0.9) + 880(1.5) = R(0.6)

P = 1765 N

F=

440(0.6)+ 880(1.5)−1335 (0.6) 0.9

F = 870 N Problem 2. Find the components in the x, y, u and v directions of the force P = 10 kN shown in the figure shown. Solution: Px = 10 cos60 = 5 kN answer Py = 10 sin60 = 8.66 kN answer Pu = 10 cos40 = 7.66 kN answer Pv=10 sin40 = 6.43 kN answer

Problem 3. The cantilever truss in the figure below is hinged at D and E. Find the force in each member.

Solution: At Joint A ΣFV=0 FABsin30∘=1000 FAB=2000N tension ΣFH=0 FAC=FABcos30∘ FAC=2000cos30∘ FAC=1732.05N compression At Joint B ΣFy=0 FBC=1000cos30∘ FBC=866.02N compression ΣFx=0 FBD=1000sin30∘+2000 FBD=2500N tension ΣFH=0 FCE=FCDcos60∘+866.02cos60∘+1732.05

FCE=2020.72cos60∘+866.02cos60∘+1732.05 FCE=3175.42N compression Summary AB = 2000 N tension AC = 1732.05 N compression BC = 866.02 N compression BD = 2500 N tension CD = 2020.72 N tension CE = 3175.42 N compression

STRENGHT OF MATERIALS Problem 1. A steel wire is 6 m long, hanging vertically supports a load of 2000 N. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 140 MPa and the total elongation is not to exceed 4 mm. E = 200,000 MPa.

Solution: Considering limitations of stress, F=

𝑃 𝐴

140 x106 =

Considering limitations of elongation, δ=

2000 N πd2 4

d = 0.00426 m d = 4.26 mm

PL AE

0.004 = πd2 4

2000(6) (200000 x 106

d = 0.0044 m d = 4.4 mm

Note: To be safe for both stress and elongation, use d = 4.4 mm

Problem 2. A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Figure. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel. Solution: Pbr = Pst= 1/(2(7848)) Pbr = 3924 N Pst= 3924 N For bronze cable: Pbr=σbr Abr 3924 = 90 Abr Abr=43.6 sq. mm

answer

For steel cable: Pst = σst Ast 3924 = 120 Ast Ast = 32.7 sq. mm

answer

Problem 3. A steel tie rod on bridge must be made to withstand a pull of 5000 lbs. Find the diameter of the rod assuming a factor of safety of 5 and ultimate stress of 64,000 psi. Solution: 𝑃

σ=k𝐴 where: k= factor of safety 5000 𝐴

64000 = (5)

A= 0. 3906 𝒊𝒏𝟐

ENGINEERING ECONOMICS Problem 1. Bessie wishes to have P 400,000 in a certain fund at the end of 8 years. How much should she invest in a fund that will pay 6% compounded continuously? GIVEN: F = P 400,000

SOLUTION:

n = 8 years

F = 𝑃𝑒 𝑟𝑛

r = 6%

400000 = 𝑃𝑒 (0.06)(8)

r= 0.06

P = P 247, 513.36

answer

Problem 2. A company invests ₱ 10,000 today to be repaid in 5 years in one lump sum at 12 % compounded annually. How much profit in present day pesos is realized? Solution: F = P (1 + i)n Where, F = Future amount P = Principal amount i = interest n = no. of months

F = 10,000 (1 + 0.12)12 F = 17,623.42 Profit = F- P Profit = 17,623.42 – 10,000 Profit = 7,623.42

Problem 3. You borrow Php. 3500.00for one year from a friend at an interest rate of 1.5% per month instead of taking a loan from a bank at a rate of 18% per year. Compare how much money you will save or lose on the transaction. Solution: Computing for the amount due after to one year. a. Borrow money from a friend 𝐹 = 𝑃(1 + 𝐼)𝑛 𝐹 = 3500(1 + 0.015)12 𝑭 = 𝟒, 𝟏𝟖𝟓

b. Borrow money from a bank 𝐹 = 𝑃(1 + 𝐼)𝑛 𝐹 = 3500(1 + 0.018)12 𝑭 = 𝟒, 𝟏𝟑𝟎 Therefore you will pay Php.55 less by borrowing the money from the bank.

STATISTICS AND PROBABILITY Problem 1. In a school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that: a. exactly 5 have access to the internet. Solution: If a pupil is selected at random and asked if he/she has an internet connection at home, the answers would be yes or no and therefore it is binomial experiment. The probability of the student answering yes is 60%=0.6. Let x be the number of students answering yes when 8 students are selected at random and asked the same question. The probability that x=5 is given by the binomial probability formula as follows: P(x=5)=8 𝐶5 (0.6)5 (1 − 0.6)3 = 𝟎. 𝟐𝟕𝟖𝟔𝟗𝟏 answer b. at least 6 students have access to the internet. Solution: P(x≥6)=P(x=6,7,8) Since all the events x=6, x=7 and x=8 are mutually exclusive, then P(x≥6)=P(x=6)+P(x=7)+P(x=8) =8 𝐶6 (0.6)6 (1 − 0.6)2 +8 𝐶7 (0.6)7 (1 − 0.6)1 +8 𝐶8 (0.6)8 (1 − 0.6)0 = 𝟎. 𝟑𝟏𝟓𝟑𝟗𝟒 answer Problem 2. In a town of 35,000 people, it is but practical to choose a representative group for a certain research project. How do you get the random sample? Use margin of error of 5%. GIVEN: N = 35,000 e = 5% = 0.05

SOLUTION: 𝑁

n = 1+𝑁𝑒 2 35,000

n = 1+(35,000)(0.05)2 35,000

n = 1+(35,000)(0.0025) n = 395.5 = 396

answer

Problem 3. Supposed a coin is flipped 3 times. What is the probability of getting two tails and one head? Solution: The sample space consists of 8 sample points. S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH} Each sample point is equally likely occurring, so the probability of getting a particular sample point is 1/8. The event “getting two tails and one head” consists of the following subset of the sample space. A = {TTH, THT, HTT} The probability of Event A is the sum of the probabilities of the sample points in A. Therefore, P(A) = 1/8 + 1/8 + 1/8 = 3/8 answer

NUEVA ECIJA UNIVERSITY OF SCIENCE AND TECHNOLOGY Cabanatuan City

COLLEGE OF ENGINEERING CIVIL ENGINEERING DEPARTMENT

APPLIED MATHEMATICS PROBLEM SET

COMPREHENSIVE EXAMINATION 1

Submitted by: Group 1 Maniego, Neal Joseph V. Dela Cruz, Nikka C. Gonzales, Anjaneth M. Nacion, Kenith A. Pagbilao, Marissa O. Tipan, Kristinne Anne D. Torio, Alyssa C.

Submitted to: Engr. MARITA DM. SANTOS Subject Professor