What about the frictional force and shear force of the bolts? two methods of thinking either the frictional resistance of the tower is 114*0.17 = 19.38 therefore to reach equilibrium the frictional resistance the combined plates create needs to equal 90.39-19.38 = 71.01kN and each bolt is...
thanks for the reply, the question is mostly aimed at bolts and threaded fasteners So I am not too worried about the forces on the cables etc I just need to get it right for my calcs. I think what they implied by saying one side broken is either front or back that way the force would be in one...
Homework Statement
Homework Equations
sum of forces in x and Y = 0
sum of the moments at a point = 0
Ffr = normal force * friction coefficient
The Attempt at a Solution
Im just trying to make a start, I am struggling to wrap my head around the basics with this question. Do you guys think...
Also we have noticed a lot of commercial motors are wrapped on angles, were not sure what or how to work out the angle required I assume it helps the motor initially turn which is one of our criteria that the motor turns without assistance at the start. Sorry if I've missed anything we are just...
Homework Statement
My group and I have to design a lap wound dc motor. Maximum of 12 volts and 5 amps. Must have no more than two poles. The aim is to achieve maximum efficiency. I've created this excel spread sheet to calculate the gauge of wire we need and the type of magnets. Also how...
Homework Statement
This is just a question relating to an assignment I am doing, if a steel block mass m1 traveling at velocity v1 and friction of 0.1 hits a concrete barrier traveling at 0 with mass m2 with friction 0.6. they have an inelastic collision and travel 0.1m together.
Homework...
Am I also correct in assuming when he is launched from the carousel, the path he travels in is a straight line perpendicular to the radius meaning the distance the kiosk is from the base shaft is the square root of the radius squared plus (the 5 metres the kiosk traveled plus the distance the...
hey thanks man that makes sense so the only forces acting on it immediately after the collision is gravity, normal force and the frictional force which means it has an acceleration of - 5.886m/s^2 over 5m which determines its initial velocity if its final velocity is 0
Homework Statement
find the velocity that the man was traveling at immediately before the collision
radius = 10m
height = 3m
angular velocity = constant
mass of person = 65kg
mass of kiosk = 150kg
displacement of kiosk after collision = 5m
coefficient of friction kiosk = 0.6
Homework...