Recent content by Corey

the question gave no indication of friction or any coefficients of kinetic or static friction so I do not think there would be any additional ground forces

"(Imagine holding something so that it just barely touches the ground. It will be touching the ground, but it's weight will be supported by your hand, so Fn would be zero. Same thing applies here, except Tsin(θ) only supports some of the weight, not all of it.)" so then wouldn't my original...

dont you have to use the torque equation to isolate the tension aspect and then solve?

I just thought it would be unnecessary because we could of isolated T already but: T=0 choosing the bottom portion of the beam as the pivot point T=0 F(x) - Tcos Theta (x)=0 Fx= Tcos theta (x) 540N (5m) = T cos (28) (3m) T=1019

Sorry about that, yes it is 28 degrees above the horizontal

A uniform vertical beam of mass 40 kg is acted on by a horizontal force of 520 N at its top and is held, in the vertical position, by a cable as shown. a) Draw a free-body diagram for the beam, clearly labeling all of the forces acting on it. b)Calculate the tension in the cable...