A storage rack to hold the paper roll. Find force reactions and draw the shear and moment diagrams for the mandrel that extends half of the roll. Use singularity functions for this problem. Givens stanchion- paper roll Density=984 kg/m^-3 OD=1.5 m ID=0.22 m L=3.23 m Density=mass/volume So W=(density*volume)*gravity Volume of hollow cylinder= (pi/4)(OD^2-ID^2)*Length of cylinder Length of mandrel=1/2 of L=>1.615 m mandrel Assumptions: The storage rack is acting with a fixed support With a distributed load of "w" where w is the weight distrubuted

A storage rack to hold the paper roll. Find force reactions and draw the shear and moment diagrams for the mandrel that extends half of the roll. Use singularity functions for this problem. Givens stanchion- paper roll Density=984 kg/m^-3 OD=1.5 m ID=0.22 m L=3.23 m Density=mass/volume So W=(densityvolume)gravity Volume of hollow cylinder= (pi/4)(OD^2-ID^2)*Length of cylinder Length of mandrel=1/2 of L=>1.615 m mandrel Assumptions: The storage rack is acting with a fixed support With a distributed load of "w" where w is the weight distrubuted

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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Answer for the 1st part of the problem. magnitude of force A = 74 N direction of force A = 25o north of west second force = P direction of force P = α south of west arrow_forward The resultant of two forces is horizontal , it means that the y component of resultant = 0 Ry = ASin(25o) - PSin(α) 0 = (74×0.42) - PSin(α) PSin(α) = 31.08 NFor, the smallest force P , the value of Sinα must be maximum which is 1 when the value of αis 90 degree Thus, P×Sin(90o) = 31.08 N P = 31.08 N The magnitude of smallest force P = 31.08 N the direction of force P = 90 degree south of west = along - y axis= arrow pointing vertically downw
%A I. Untitled Section F1 F2 a C F3 A b a F4 D E B Consider the following values: - F5 a = 16 m; b = 11 m; c = 13 m; d = 8 m; F1 - 3 kN; F2 - 11 kN; F3 - 2 kN; F4 - 20 kN; F5 - 2 KN; a = 30°, 0 = 60° , B= 30° 11 What is the resultant moment of the five forces acting on the rod about point A? a) 88.8 kN.m b) 76.3 kN.m c) 19.6 kN.m d) 55.5 kN.m e) 59.1 kN.m f) 81.3 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 88.8 kN.m b) 120.8 kN.m c) 31.6 kN.m d) 98.9 kN.m e) 41.1 kN.m f) 24.2 kN.m 31 What is the resultant moment of the five forces acting on the rod about point C? a) 125.7 kN.m b) 178.1 kN.m c) 254.6 kN.m d) 439.1 kN.m e) 144.1 kN.m ) 215.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 77.8 kN.m b) 98.3 kN.m c) 100.6 kN.m d) 18.9 kN.m e) 97.1 kN.m H 60.1 kN.m 5] What is the moment of the force F2 about point E? c) 66.6 kN.m Activate W f) 151.3 kN.m a) 122.7 kN.m b) 108.1 kN.m d) 83.1 kN.m e)…
A gymnastic beam can be modeled with a 10 cm x 40 cm cross-section and 6 m between its supports. A gymnast lands at the center of this beam with a 1.5 kN force. What is the shear stress on the upper surface of the beam directly below the force of the gymnast?
A sandwich board advertising sign is constructed as shown in the figure below. Hingen 0.50 m Chain 1.30m CG CG Uniform board (cG at center) 1.10 m The sign's mass is 7.20 kg. (a) Calculate the tension in newtons in the chain assuming no friction between the legs and the sidewalk. (b) What force is exerted in newtons by each side on the hinge?
W1 weight is dropped from height h towards W2 weight,W1=40N,W2=400N,h=200mm, k=32 kN/m.Find the maximum values of the spring force and the bending amount of W2.Consider the impact of between W1 and W2 is not flexible,it's rigid, ignore the weight of spring,and solve it using energy conversation method. I am Turkish,if my english is bad I am so sorry I did my best.Can you help me about this problem?
Calculate the reaction forces on the bearing near the pulley and bearing at the far end of the shaft. Include a free body diagram and coordinates. Given the shaft length is 1 meter= Mass=40Kg Static load = 8600N Dynamic (running load) = 8400N Some hints:1. This solution will require you to draw known forces and find unknown reactions in 2perpendicular planes and then add the horizontal and vertical reaction forces asvectors to find the reaction force (hypotenuse).2. Bearings are usually mounted as near as possible to the ends of the shaft withoutfouling the other machine elements. The bearing at the far shaft end can be mountedat the 1 meter point. The bearing at the pulley end should allow clearance to assemblethe pulley.3. Bearing reaction forces are typically simplified as a single point load through thebearing centre.
ENTER YOUR ANSWER THE WAY I SPECIFY HERE IN ADDITION TO DOING IT ON SCRATCH PAPER 400 N 800 N B D A E C 2 m Use the method of sections to calculate the axial force in members AC and BC. Enter your answer in the following way rounded to the nearest whole number. Example If your answer is TAC 2 N tension TBC =3.1 N compression. enter 2Tension;3Compression Enter TAC then TBC (no space)
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.)
Suppose beam of the list on a comer A and against a wall at the point B( No friction any surfaces) The Beam angle 30 degrees. The force between the beam and the Conner (at point A) is 10N, Calculate X Component of force this force given the coordinate axes defined in the Figure. Suppose a beam of lengh L sts ona comer Aend againat a wal at point 8 (Aume na tion on any sufacen). The bean mtnde degrees with the comer e shown in the gun Asune he force betenen the bean and the comer a pont Ajis 10N, cae ha bron given the coordeatn auna dafred in the fgum II. Calculate all the Support reaction at B. For the purposes of Submission Divide all your force Values by q0*a (ie. Fa/(q0*a), thereby yielding a unitless number 90 D E 2a B.
The pipe is pulled by three cables as shown on the figure below. All three cables are subjected to known forces. Another cable is planned to be added on the system. This additional cable is designed to pull the pipe along the positive y-axis. The proposed cable, Fy is twice as much as the only force which tends to pull the pipe below the x-axis. (Fa= 7.5kN; Fb= 1.25kN; Fc=5kN) Find the summation of forces in x axis and y axis.
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1. - solve the nodal displacements and reaction forces: Al 20 kN/m 20 kN/m 20 kN/m 20 kN/m 5 kN ww Nom