Recent content by paolostinz

Wow, thank you, I can't believe I missed that part about the friction, I just assumed it was acting downwards. Alright this is my second attempt: F_a=1.20 x 10^4 N [up] F_g=mg F_g=(1030 kg)(9.8 m/s^2 [down]) F_g=1.01 x 10^4 N [down] F_net=F_a + F_g F_net=1.20 x 10^4 N [up] + 1.01 x 10^4 N...

Homework Statement An elevator that contains three passengers with masses of 72 kg, 84 kg, and 35 kg respectively has a combined mass of 1030 kg. The cable attached to the elevator exerts an upward force of 1.20 x 10^4 N, but friction opposing the motion of the elevator is 1.40 x 10^3 N...

You both are/were a tremendous help, thank you!

In the link you provided it says that north or south should be stated first always. Could you elaborate a little bit because this part has confused me in the past? I usually just go with my gut which isn't entirely accurate.

Yes, that is what I meant. I'm not sure if that's the correct way to reference it, it's just the only way I thought of describing it. Thank you so much for your help! Btw, how do you attach thumbnail like that?

Sorry lol, let me clarify, they are meant to indicate direction; [north 10 degrees west] in reference to the log's original position.

So what you're saying is I determined the wrong angle? Or did I even need to determine the angle? Should my answer be 1164 N [N 20 W] ?

Homework Statement Two tugboats are pulling a large log, as shown in the following diagram. The log has a mass of 250 kg and is initially at rest. How far have the tugboats moved the log after 10 s? http://imgur.com/GS7Y80x Homework Equations c^2=a^2 + b^2 -2ab cosC sin A/ a= sin...

See, that makes sense, it's the little things like this that confuse me most. I wish I had better access to teachers to clear up these type of hiccups. I'm so grateful for this place, thank you everyone!

Yes, thank you! I see now, it seemed like I was getting the wrong answer when I was doing the original equation, so I thought something else was at play here. Is there any reason why one would prefer writing the equation this way? I just don't get why my textbook would switch the equation...

My online physics course is using cosine law to find the net forces on objects. My question is to do with the equation, at first it shows it as : c^2=a^2+b^2-2ab cosC. From there, it changes to: c=[a^2+b^2-2ab cosC]^1/2. How and why does this work? Why isn't a square root involved in the...

Wait, you're not SteamKing, I mistook you for him, but I will say a thanks to you as well mic* !

The way my book has been showing me it seems to round a lot of the time, specifically in cases such as this one. I too agree that it's a substantial difference, but I also don't want my answers to deviate too far from their answers/ how the textbook has taught me. I assume that it's just because...

Alright, I'm with you so far, I hope; Here's my new attempt: d_v=1/2at^2 -150=-4.9 t^2 t=6 s d_h=vt d_h=(40 m/s)(6 s) d_h= 240 m

So then should I be trying to find the initial velocity then? I'm not given any angles do I don't see how to go about doing that. My intuition tells me that I should be determining time but my answers seem wrong. Any advice?