IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: The force of 600 N is the resultant of the forces F1 and 240 N acting as shown in the figure. Which of the following gives the value of F1. 600 N F1 3 4 12 5
240 N
Choices: a. 698.9223 N, 56.310 with respect to the horizontal b. 698.9223 N, 33.690 with respect to the horizontal c. 698.9223 N, 56.310 with respect to the vertical d. 698.9223 N along the horizontal ANSWER: A SOLUTION ( IF PROBLEM -SOLVING) F1X + 240(5/13) = 600 (4/5), F1X = 387.6923 N F1Y – 240(12/13) = 600(3/5), F1Y = 581.5385 N F12 = (387.6923)2 + (581.5385)2, F1 = 698.9223 N θ = tan-1(581.5385/387.6923) = 56.310 Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 63 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: Given the set of parallel forces shown, which of the following gives the resultant force. 100 KN 40 KN 60 KN 50 KN 60 KN A
B 2m
2m
Choices: a. 90 KN up, 3.33m fr A c. 90 KN up, 3.33m fr B ANSWER: B
2m
6m
3m
b. 90 KN down, 3.33m fr A d. 90 KN down, 3.33m fr B
SOLUTION ( IF PROBLEM -SOLVING) R = 60 + 50 -100 -40 -60 = -90 KN ΣMA = Rd, 100(2) + 40(4) – 60(6) – 50(12) + 60(15) = 90d, d = 3.33m Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 63 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: The coefficient of friction between the rope and fixed post is 0.30. The rope is wrapped only once around the post. Which of the following gives the minimum tension in the rope at pt A A
B 300 N Choices: a. 1024.7681 N b. 7936.1959 N c. 138.6873 N ANSWER: C SOLUTION ( IF PROBLEM -SOLVING)
d. 69.3437 N
300/T = e0.4Π, T = 138.6873 N Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 179 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: The resultant of the concurrent forces has a magnitude of 1200 N and acts through the origin and points x = 3m, y = 4m, z = 5m. Which of the following gives the x component of the resultant force Choices: a. 678.8194 N b. 848.5243 N c. 240 N ANSWER: D SOLUTION ( IF PROBLEM -SOLVING) d2 = 32 + 42 -52, d = 7.0711 m Fx = 1200(3)/7.0711 = 509.1146 N
d. 509.1146 N
Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 28 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: From the speed of 100 kph, a car decelerates at the rate of 15 m/min/sec along a straight road. Which of the following gives the distance travelled at the end of 40 sec. Choices: a. 3800 m b. 911.112 m c. 91.111 m ANSWER: B SOLUTION ( IF PROBLEM -SOLVING) V0 = 100kph x 1000 /3600 = 27.7778 m/s a = -15/60 mps2 = -0.25 mps2 s = 27.7778(40) + 0.5(-0.25)(40)2 = 911.112 m
d. 455.56 m
Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 395 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: The velocity of a particle moving along the x axis is defined by v = x4 + 2x3 – 3x2 where v is in m/s and x is in m. Which of the following gives the velocity when x = 2m. Choices: a. 20 m/s b. 44 m/s c. 40 m/s ANSWER: A SOLUTION ( IF PROBLEM -SOLVING) v = 24 + 2(2)3 – 3(2)2 = 20 m/s
d. 22 m/s
Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 394 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: The block shown reaches a velocity of 10 m/s in 25 m starting from rest. Which of the following gives the coefficient of kinetic friction between the block and the ground. 600 N Vo = 0
V1 = 10 m/s 280 N 25 m
Choices: a. 0.67 b. 0.67 N c. 0.2626 N ANSWER: D SOLUTION ( IF PROBLEM -SOLVING) 102 = 2a(25), a = 2 m/s2 ΣFx = max, 280 – μ600 = (600/9.8)2, µ = 0.2626
d. 0.2626
Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 500 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: The normal acceleration of a particle on the rim of a pulley 10 ft in diameter is constant at 1200 fps2. Which of the following gives the speed of the pulley in rpm. Choices: a. 77.4597 b. 15.4919 c. 147.9371 d. 73.9686 ANSWER: C SOLUTION ( IF PROBLEM -SOLVING) an = v2/r, 1200 = v2/5, v = 77.4597 fps v = rω, 77.4597 = 5ω, ω = 15.4919 rad/sec*60/2Π = 147.9371 rpm Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 425 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: A 350 N block is resting on a plane inclined at 300 with the horizontal is subjected to a constant force of 400 N acting parallel to the inclined plane. After the block has moved 3 m from rest, the 400 N force was removed. Coefficient of friction is 0.2. Which of the following gives the distance that the block will move further along the inclined plane until it stop. 350 N 400 N
300 Choices: a. 2.714m b. 3m c. 5.714m d. 5.428m ANSWER: A SOLUTION ( IF PROBLEM -SOLVING) WE equation: ΣFs = 0.5m(v2 – v02) 400(3) -0.2(350sin 30)(3+x)-(350sin 30)(3+x) = 0.5 (350/9.8)(0-0) X = 2.714 m Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 536 Endorsed by: School Representative
IECEP Manila
IECEP-Manila Form 1 S.2011
TEST QUESTIONS
1. Please Use One Page per Question. 2. Fill-out all needed information. 3. Submissions with incomplete data will not be considered. PROFILE OF CONTRIBUTOR Name of School UNIVERSITY OF SANTO TOMAS Address Espana Blvd., Manila Signature over Printed Name , Position Ma. Luisa T. Asilo, Faculty Area of Specialization
Statics and Dynamics of Rigid Bodies
Please indicate the subject: (x) General Engineering and Applied Science ( ) Electronics System & Technologies QUESTION: The force system consist of a clockwise couple of 600 N-m plus a 300 N force directed to the right through the origin of x & y axes at θx = 300. 300 N 600 N-m
0
30
O If the system is replaced by an equivalent single force, which of the following is its x and y intercept. Choices: a. 4 m right and 2.3094 above the origin b. 4 m left and 2.3094 above the origin c. 4 m left and 2.3094 below the origin d. 4 m right and 2.3094 below the origin ANSWER:C SOLUTION ( IF PROBLEM -SOLVING) sin 30 = 2/x, x = 4m left of origin tan 30 = y/x, y = 2.3094m above the origin Reference: Engineering Mechanics 3rd Edition by Ferdinand L. Singer page 64 Endorsed by: School Representative
IECEP Manila
