Recent content by gv3

Is that because the shoe is keeping the ladder up and also preventing it from sliding outwards?

So when you draw the fbd, the force from the rail is on an angle and the two forces from the shoe are vertical. Doing it this way you wouldn't get a force of 294.5N from the shoes. but a force of 294.5N is what gives the right answer.

Does the other factor of two come from the ladder shoe? since there's two points of contact from the ladder shoe onto the bolt? Since the shear force is parallel to the ladder is it just the resultant force i found in part 2? Ok so going with what i said above i got the correct answer. I did a...

I thought shear forced always occurred in the Y direction. When drawing my fbd i have V pointing in the negative Y direction and Ay pointing positively in the y direction. maybe i missed some forces from the ladder shoe? I am not sure where the factor of 4 comes from. is it from the ladder rail...

Homework Statement [/B] The inclined ladder AB supports a house painter (85 kg) at C and the self weight (q = 40 N/m) of the ladder itself. Each ladder rail (tr = 4 mm) is supported by a shoe (ts = 5 mm) which is attached to the ladder rail by a bolt of diameter dp =8 mm. 1) Find support...

yeah it would just be the position vector divided by the magnitude.

My mistake you are right. How would the velocity vector be found then? My professor in his lecture notes looks like he just took the direction of V shown.

But isn't that vector 2i-4j-4k?

1. Homework Statement Determine the total magnitude of angular momentum Ho of the particle about point O. The velocity of the particle is 5.5 m/s.Homework Equations Ho= r x mv The Attempt at a Solution The answer is 43.04. My question is, isn't the graph wrong? If you take the magnitude of the...

The next question asked for the max stress/average stress and used 30ksi as he average stress which gave me the right answer. I am hesitant to try because i only have 2 attempts left.

Not sure. I've still been unable to get the answer. I don't see any other way to approach this. My professor approached it the same way when he did a similar problem.

well because i was thinking that not only will the rail in the middle expand, but the one on the left and right of it would also expand. So if the deformation wasn't split between both ends, the gap would have to be 2δ. Because the rail in the middle will expand by δ and the rails next to it...

Homework Statement The stress distribution along the section of a bar is shown below. From this distribution, find the approximate axial force P in kip. Homework Equations δavg=P/A A= (.6+.6)(.5) δavg=30ksi[/B]The Attempt at a Solution P=(30ksi)(.6in2)=18kip It says this answer is wrong...

Homework Statement The 40 ft long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determine the required gap in inches so that the rails just touch one another when the temperature is increased from -20 F to 90 F. The cross sectional...

Thanks for helping! :D