QUESTION 8The y-components results of the force vectors that has been shown in Figure.F2 = 80 N20%3DF = 150 N%3D30°15°F = 100 NAF3 = 110 N%3DFly=75 N, F2y=-75.2 N, F3y=-110OA."N, F4y=-25.9 NFly=75 N, F2y=75.2 N, F3y=110 N,OB.'F4y=-25.9 NFly=75 N, F2y-75.2 N, F3y=-110'N, F4y-25.9 NFly=-75 N, F2y-75.2 N, F3y=-110'N, F4y=-25.9 ÑOC.OD.

To find: The y component results of the force vectors. Concept used: Compute the force in…

Answered: F2 = 80 N 20° F = 150 N 30° A 15° F =…

Engineering

Mechanical Engineering

F2 = 80 N 20° F = 150 N 30° A 15° F = 100 N %3D F3 = 110 N %3D

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

See similar textbooks

Similar questions

The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. d 2 Pl CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m С. 62,5 N*m D. Object 1 doesn't exert a torque.
Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. 01 = 35 degree and 02 = 03 = 20 degree. Calculate the %3D moment that the force, F, exerts .about the point A F A. M = -1042 O M = -1055 M = -1046 M = -1050 O
Figure F= Value Submit Request Answer Units A < Return to Assignment Provide Feedback 4 m 30° Determine the magnitude of the projection of force F=680 N along the u axis (Eigure 1) Express your answer to three significant figures and include the appropriate units. U1 4 m 2m < -y F 1 of 1
Consider the bracket shown in Fig-1. If: F1 = 10 N@ 120° • F2 = 75 N@ -90° then, what is the magnitude of the resultant force on the bracket? 8.66 N 66.5 N -75 N 75 N O ON 66.3 N*m 100 N*m O 10 N
The figure below shows a steel structure made up of two pre-assembled segments that are hoisted by a mobile crane using a bespoke lifting frame. Each cable must withstand loads of FAB = FAC = FAD = FAE = 4000 kN. You must provide an accurate answer in scientific standard form with one decimal place only. In cartesian vector and in scalar form, compute the rAB, rAC, TAD & rAE length. 14 m 6m 6 m в 3 m 3 m X
Engineering Analysis ENGR 1022. Vectors Homework #3 1. Compute the dot product for the following two vectors. A=51+3) and = 31-10) 2. Compute the cross product for the following vectors? C= 61-9) and 5 =91-4k 3. What is the force in cable AC? See diagram below. B 500 lb 4. The magnitude of torque equals force times distance. The vector form is r=7x F. F is the force vector and F is the position vector. You are pushing on a merry-go-round with a radius of 5 ft with a force of 40 lbs. What is the torque vector value? The z axis is pointing out of the sheet of paper. See diagram below. y 5. Show that vectors 7,7, and F in problem 4 are all perpendicular to each other. University of Arkans Fot S5210 Grand Avenue PO, Box 3649 Fort South, Arkama 72933-3649-475-753-7000
a = 1 m b = 2 m c = 3 m Force AB = 500 Newtons FAB Find the y-component of unit vector AC. (Hint: the value should be less than 1). Find the z-component of unit vector AC. (Hint: the value should be less than 1). Find the x-component of Force AB in N. If the calculated answer is not whole number, express it in 4 significant figures. If the value is negative, include a negative sign.
3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)
The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point P
Example ( 2-1) : The ring in the figure below is subjected to two forces ( F1) and ( F2). Determine the magnitude and direction of the resultant force. 10° F, = 150 N F = 100 N %3D 15° Solution : Parallelogram Law. The parallelogram is formed by drawing a line from the head of F1 that is parallel to F2 , and another line from the head of F2 that is parallel to F1 , the resultant force FR extends to where these lines intersect at point A. The two unknowns are the magnitude of ( FR) and the angle ( 0) (theta).
Assign the values of shear force and bending moment to the image bar as a function of the x coordinate. q=20kN/m F=14kN M=12kNm L1=1.2m L2=1.4m L3=1.3m L4=1.5m Minimum value of shear force: Qmin = Maximum value of shear force: Qmax= Use units (kN,m), Do not round the invoices in the intermediate stages, but only the final answer!
S00 Ib In the fig. shown, compute the ff: (16-18) the resultant using cosine law (force polygon) 60 R = 35 (19-20) the angle of the R measured 500 lb cW from the x- axis.