piper210_355 Posted October 9, 2008 Share Posted October 9, 2008 Can someone please help me. My teacher hasn't taught us anything, yet she expects us to be able to do this. A 2.00 kg box is at rest on a horizontal floor. A horizontal force of 18.0 N is required to keep the 2.00 kg box moving across the floor at constant speed. A box of unknown mass is stacked on top of the 2.00 kg box. A horizontal force of 56.0 N applied to the 2.00 kg box is required to keep both boxes moving across the floor at constant speed. What is the mass of unknown box? Link to comment Share on other sites More sharing options...
Klaynos Posted October 9, 2008 Share Posted October 9, 2008 So, what do you know about friction forces? What equations? Link to comment Share on other sites More sharing options...
piper210_355 Posted October 9, 2008 Author Share Posted October 9, 2008 I know that F(friction)=mF(normal) but i don't know how to find these I know that normal force is the weight of the object in this case, since it is parallel to the ground. Link to comment Share on other sites More sharing options...
Klaynos Posted October 9, 2008 Share Posted October 9, 2008 What is the normal force equal to? (I assume m is the coefficient of friction?) Link to comment Share on other sites More sharing options...
piper210_355 Posted October 9, 2008 Author Share Posted October 9, 2008 yes that's the way it's written in my book. Link to comment Share on other sites More sharing options...
mooeypoo Posted October 10, 2008 Share Posted October 10, 2008 piper, in questions like these, the first thing that will help you is draw a visual scheme of the physical situation. Draw the body and all the forces that apply on it in all directions. Looking at it will help you know what force applies to what, and which forces equal what.. (big hint here If you want, post your drawing here, and we'll help you make sense of it. ~moo Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 ok i looked over my physics book and i think i solved it i used this equation to get the static friction: u=F(s),max/(mass)(gravity) i did it for the 2kg box when it was being pulled first. U(s)=18/(2)(9.81) I got .92. The next part is what i don't know if you can do or not i used the same equation for when both boxes are being pushed, and i put .92 as the static friction for that (although i'm not sure if i can do that). .92= 56/(2x)(9.81) I got 3.10kg Link to comment Share on other sites More sharing options...
big314mp Posted October 12, 2008 Share Posted October 12, 2008 You have the right idea, but you should just use (x+2) instead of 2x. If you use 2x, you are assuming that both boxes weigh the same, when in fact one weighs 2kg and the other is unknown. Link to comment Share on other sites More sharing options...
piper210_355 Posted October 12, 2008 Author Share Posted October 12, 2008 that make sense. i did it again with (x+2) and got 4.20. Thank you sooooooo much for your help. Link to comment Share on other sites More sharing options...
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