QUESTIONS 2.1. A force of P defined by the angle 0, = 75° to the horizontal acts through a point. What are the components of this force on the x- and y-axes? [See Figure 2.59(Q1).] <= 0.97P. Answer: P 0.26P, P, = OFOR and 150P spectively
QUESTIONS 2.1. A force of P defined by the angle 0, = 75° to the horizontal acts through a point. What are the components of this force on the x- and y-axes? [See Figure 2.59(Q1).] <= 0.97P. Answer: P 0.26P, P, = OFOR and 150P spectively
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![2.6 times that developed in the aluminum: f, 2.6f, Because P AJ Apanhola
+ A,J. (2.6) (0.5) () + (0.5) () 5000, we have f. 2777 lb/in. and f-
7220 lb/in.¹ (The nA, value is often called a transformed area; the steel is considered to have
been transformed into a modified area of aluminum that is structurally equivalent.) To find fal
deformations, fa/B. 2777/11,300,000 0.000246 and AL, L-0.00245 in
=/ft
ANA
QUESTIONS
2.1. A force of P defined by the angle 0,= 75° to the horizontal acts through a point. What
are the components of this force on the x- and y-axes? [See Figure 2.59(Q1).]
Questions.
211-
L/2-
2.2. The components of a force on the x- and y-axes are 0.50P and 1.50P, respectively.
What are the magnitude and direction of the force? [See Figure 2.59(Q2).J
Answer: 1.58P at 0,= 71.5°.
2.3. The following three forces act concurrently through a point: a force P acting to the
right at 0, 30° to the horizontal, a force P acting to the right at 0,= 45° to the hori-
zontal, and a force P acting to the right at 0,= 60° to the horizontal. Find the single
resultant force that is equivalent to this three-force system. [See Figure 2.59(Q3).]
Answer: 2.93P at 45°.
+5A+
Answer: P = 0.26P, P,= 0.97P.
Py
1.5P
L
2P
(Q2)
W
FitiFit font
(Q6)
FL/3+2L/3-
(Q10)
Peo
P 2000 lb
(Q13b)
5 kips
(Q3)
15 ft
(Q7)
441441 -
(Q11)
P45
W= 2 kips
4 ft 4 ft 12 ft
P30²
(Q13c)
2P
P
P
(04)
4P
2P
APĮ I
(Q8)
W
NNN
(Q12)
8 = 45°
+12+12 +212 +212 +2/2+1/2-
(Q13d)
45°](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa24caf9c-9cb0-4d81-9388-d80f1be9da2e%2F811216bd-f612-447f-b660-304ee5310bdc%2F37yt8k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2.6 times that developed in the aluminum: f, 2.6f, Because P AJ Apanhola
+ A,J. (2.6) (0.5) () + (0.5) () 5000, we have f. 2777 lb/in. and f-
7220 lb/in.¹ (The nA, value is often called a transformed area; the steel is considered to have
been transformed into a modified area of aluminum that is structurally equivalent.) To find fal
deformations, fa/B. 2777/11,300,000 0.000246 and AL, L-0.00245 in
=/ft
ANA
QUESTIONS
2.1. A force of P defined by the angle 0,= 75° to the horizontal acts through a point. What
are the components of this force on the x- and y-axes? [See Figure 2.59(Q1).]
Questions.
211-
L/2-
2.2. The components of a force on the x- and y-axes are 0.50P and 1.50P, respectively.
What are the magnitude and direction of the force? [See Figure 2.59(Q2).J
Answer: 1.58P at 0,= 71.5°.
2.3. The following three forces act concurrently through a point: a force P acting to the
right at 0, 30° to the horizontal, a force P acting to the right at 0,= 45° to the hori-
zontal, and a force P acting to the right at 0,= 60° to the horizontal. Find the single
resultant force that is equivalent to this three-force system. [See Figure 2.59(Q3).]
Answer: 2.93P at 45°.
+5A+
Answer: P = 0.26P, P,= 0.97P.
Py
1.5P
L
2P
(Q2)
W
FitiFit font
(Q6)
FL/3+2L/3-
(Q10)
Peo
P 2000 lb
(Q13b)
5 kips
(Q3)
15 ft
(Q7)
441441 -
(Q11)
P45
W= 2 kips
4 ft 4 ft 12 ft
P30²
(Q13c)
2P
P
P
(04)
4P
2P
APĮ I
(Q8)
W
NNN
(Q12)
8 = 45°
+12+12 +212 +212 +2/2+1/2-
(Q13d)
45°
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