Recent content by hollidca

Not only wider but no ridges, so it is smooth and therefore more road contact. Well for your answer I'd quote the formula, explain how surface area isn't a factor, but then counter argue and say how this formula is only an approximation and surface area does matter. Then write a short...

Fr = μN where: * Fr is the resistive force of friction * μ is the coefficient of friction for the two surfaces (Greek letter "mu") * N is the normal or perpendicular force pushing the two objects together * μN is μ times N surface area doesn't come into it.

I think you'll find this very useful, http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Friction.html

Yes, It's a tower crane. We are assuming zero friction and also that there are no impact loads such as wind, snatching, or just impact of another object. Typical undergraduate course lol, always assuming things are insignificant when they are the most significant.

Okay, thank you. I think i have sorted it out in my head. Now to do it on paper... The joys of force analysis.

No, 1.3x10^5 is pressure in chamber 2. Use the forumla given to you in the question to calcualte the pressure in chamber 1. You know the area of chamber 1 and you have calculated the pressure for chamber 1 so rearrange the formula pressure=force/area to give you the force.

In ME we have an entire module for 3 or more years based around renewable energies, fossil fuels, fluid dynamics, thermal dynamics and the like. So you're still covered for gas/oil etc. Also because we get an far wider understanding of other engineering disciplines we are more likely to get hire...

Well the good news is most university engineering courses share the first year, or most of it anyway. Personally I am a ME and i LOVE it! Can't stand the electrical side of it, apart from logic, then again I'm not a fan of maths. I suggest you get in touch with your uni and discuss the options...

Okay find out the pressure in chamber 1. Pressure=force/area. Did this help?

Okay I think i get what your saying. Consider a 10N Load pulling to the right around the pulley, and the opposite 10N pulling down. Leaving a resultant at 45 degrees from the horizontal and vertical of 10 root 2. This would need an equal and opposite force? Meaning that the beam that the pulley...

Hi for my design project at uni I am to design a truss crane to a certain spec. I can easily calculate the forces in a truss without the pulley systems. However when I have the pulleys I'm not sure how this affects the forces in the truss. I have the load attached via a 4 cable pulley at the...