Recent content by nialltm1991

many thanks that really helps me. so I would then use the formula shear stress= load/area? so in that case mild steel has a shear strength of 80kN/mm2. So the load required to strip the thread would be 80,000=load/(4xL) where L is length of the nut? the object of the report is to prove that...

Hi all, im building a bench vice for college and I need to do a report on It also. the problem I am having is working out the force required to strip the threads off an M10 x 1.5 metric thread. both the thread and nut are to be made from mild steel. any info that I could find online has only...

forces up = forces down forces left = forces right Rn + 300sin(theta) = mg F = 300cos(theta) Rn + 300sin(theta) = 500.31 muRn = 300cos(theta) because F = muRn Rn = 300cos(theta)/mu...

yes i can i have a fbd ... but then i will have 2 equations with sin in one and cos in the other and that is the part i get stuck at

Homework Statement im having problems with this question...a body weighing 51kg is pulled on a horrizontal plane with coefficient of friction 0.3. a force of 300N is applied at angle x ... find the angle x? Homework Equations The Attempt at a Solution the only solution i have come...