Concept explainers
(a)
Answer to Problem 13.82P
Explanation of Solution
Given:
(b)
That the work done by
Answer to Problem 13.82P
Work done by
Explanation of Solution
Given:
O − A :
Also, on O − A
And
Thus,
A − B :
Also, on A − B
And,
Thus,
B − D :
Also, on B − D
And
Thus,
Thus,
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Chapter 13 Solutions
Vector Mechanics For Engineers
- Force P with magnitude of 15 kN is acting at B (-1,2, 6) m to C (5,4, 4) m. Force Q with magnitude of 37 kN is acting at D (1,-5,3) m to E (-6,1, 5) m. Force R with magnitude of 28 kN is acting at A (5,2, 7) m to F(-8,1,8) m. Determine the equivalent resultant force and couple moment at point O (4,-5,6) m.arrow_forwardВ Бm 5 m 5 m 5 m 5 m А 5 m C 5 m E 30 kN 20 kN The magnitude of the moments about A induced by internal force AC is kN*m.arrow_forwardThe moment of the force F=360-lb about the point O in Cartesian vector form is. 1 ft 4 ft 2 ft 4 ft Select one: a. Mo = 600j- 1200k O b. Mo=-200j+ 400%arrow_forward
- Given the Position vector R = i+0j + k (m) and the Force vector F = 2i + 2j + 2k (kN) Determine R. F Answer:arrow_forwardA force F is represented by the vector 3i+4j+5k. Find the component of force F acting perpendicular to a plane, thi, which contains the vectors P=i-j, and Q=i+j-k (P and Q lie on thi).arrow_forwardCalculate the work required to move a mass “m” against A Force Field F = 5Ux + 7Uy along the indicated direct path from Point A to Point Barrow_forward
- The resultant of the distributed load in the section between A and B is equal: 1.5 kN/m A В 3 m R=4.5 kN at distance 2m from A R=2.25 kN at distance 2m from B R=2.25 kN at distance 2m from A R=2.25 kN at distance 1.5m from Aarrow_forwardThe moment of the force F=120-lb about the point O in Cartesian vector form is. F-120 lb 1ft 4 ft 2 ft Select one: a. Mo = 200j - 400k b. Mo = -200j+ 400k c. Mo = -200j- 400k O d. Mo = 200j+ 400karrow_forwardQ4\ The location of the resultant force of the system below with respect to point C is: 200 #3 #3 360 3. # 2 200 %23 220arrow_forward
- F1 F2 F3 A B b F4 E Consider the following values: - F5 a - 25 m; b - 5 m; a - 25 m; d- 5 m; F1 - 4 kN; F2 2 kN; F3 - 5 KN; F4 - 7 kN; F5 - 3 kN; = 30°, 0 = 60° B= 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 4.5 KN.m b) 1.7 kN.m ) - 82.6 KN.m d) 82.9 kN.m e) 24.1 kN.m 0- 47.1 KN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) - 96.1 kN.m b) 21.7 kN.m e) - 41.6 kN.m d) 80.9 kN.m e) - 5.4 kN.m D 87.2 KN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) - 98.6 kN.m b) 66.1 kN.m ) 17.6 KN.m d) 28.9 KN.m e) 91.1 kN.m ) 22.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) – 109.9 KN.m b) 112.7 KN.m )-81.9 kN.m d) 126.9 kN.m e) - 127.1 kN.m 030.1 kN.m 5] What is the moment of the force F2 about point E? a) 39.6 kN.m b) 48.1 kN.m c) 25.8 kN.m d) 13.1 kN.m e) 11.1 kN.m ) 21.3 KN.marrow_forwardThe two forces are equivalent to a force that has a line of action passing through point A y 0.2 m- 120 N 490 N 0.1 m A None of these O R=150 N, x-0.075 m R=150 N, x-0.1 m R=150 N, x-0.133 m O إرسال رجوع عدم إرسال كلمات المرور عبر نماذج Go ogle مطلقًا۔ eENVI ALMansOur University College l: ziill lia al :Larrow_forward3. Find the resultant force and its direction. Use the Ccmponent method. 100 N 110N 30 3ON 160Narrow_forward
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