Resolveforce F1 andF2 into theircomponentsacting alongu and v.70°45°F₂= 500 N30°F₁= 300 N

According to the given dataHere F1=300N and F2=300NWe need to resolve these forces along the u and v

Answered: Resolve force F1 and F2 into their…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Q1) Determine the magnitude and direction of F3 so that If the resultant of the three forces act on the bracket has a magnitude of 250 N and a direction of 20° measured clockwise from the positive x axis. F₁ = 100 N 13 12 2 ME 0 F2 = 150 N F3 X
If the coordinate direction angles for F3 are α=120°, ³³=60° and y3=45°, determine the magnitude and coordinate direction angles of the resultant force acting on the eyebolt. (Ans: 1.55 kip, 82.4°, 37.6° and 53.4°). F₁ = 700 lb 1 F3 = 800 lb F₂= = 600 lb
The magnitude of the component of force F2 acting on the hook with the X axis equals to; 30° 30° F1 = 300 N F2 = 500 N
30° 75° F = 4 kN 30° F2 = 6 kN Resolve force F 2 into component vectors acting along directions u and v. Determine magnitudes of each direction. Use Law of Sines and Cosines.
Find the magnitude of the force F1 and its direction 0 given that the resultant force is 430 N and directed along the positive y axis. Set F2= 860 N F1 15° F2 1056 N, 0= 51.85 deg. 856.2 N, 0= 38.15 deg. 856.2 N, 0= 77.84 deg. 1056 N, 0= 38.15 deg. 1056 N, 0= 77.84 deg. 856.2 N, 0= 51.85 deg.
An object is subjected to four forces as shown in figure Q1(a). Determine the magnitude and direction of the resultant force. F2= 600 N F = 800 N 45°- 60 F4= 300 N 13 12 F = 650 N
The 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 :L
Express the 40 lb force in the rope from point A to point C in cartesian format. Point C is in the x-y plane and at an angle of 20 degrees from the y axis as shown. 4 ft 20 lb A B 3 ft X 40 lb Z 2 ft 1.5 ft 4.87i+27.6j - 28.5k lb -4.87i+27.6j + 28.5k lb -4.87i+27.6j - 28.5k lb 4.87i+22.8j - 32.5k lb 2 ft 20 y C
A 44 lb force is directed along a cable that goes from the origin to point A . What is the x component of the force? N y F ✔
F2-13. Determine its coordinate direction angles of the force. 45 -30° F = 75 lb F2-13 F2-14. Express the force as a Cartesian vector. F = 500 N 60° 60° F2-14 F2-15. Express the force as a Cartesian vector. F = 500 N 60 F2-15 F2-16. Express the force as a Cartesian vector. F-50 lb 1445 F2-16 F2-17. Express the force as a Cartesian vector. F = 750 N 45% F2-17 F2-18. Determine the resultant force acting on the hook. F₁ = 500 lb 30 45° F₂=800 lb F2-18 23
Determine the magnitude and coordinate direction angles of the resultant of F₁, F2, and F3. |FR|= ex(x) = 0,(B) = 0₂(y) = F₁ = 200 N 30° 30° 4 m 45° F₂ = 180 N 60° 2 m y 1.5 m F₁ = 320 N

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Resolve force F1 and F2 into their components acting along u and v. 70° 45° F₂= 500 N V 30° F₁ = 300 N