Recent content by MGCLO

Normal Force is equal to the force acting on the surface. which is in case Y-Component of G.and it must be perpendicular to the surface. so when we have an angle over here. you need to get the Y-component of the FG(Force of Gravity). which is equal to the normal force. if you don't have an...

B. because on the outside. there is more fluid. therefore. more weight that is the fluid putting on the glass. but on the inside. less fluid and the same surface are touching the tube! so it would be easier for the fluid to rise inside the tube. this question is really common sense! i never did...

when i do the geometry i get an unreasonable answer of cosθ= adjacent/hyp cos10= 50/ hyp hyp=50/ cos10 hyp = 49.24 N. is that correct? i did it wrong last time. i believe this is correct. answer is reasonable this time!

here is how the problem looks. http://i39.tinypic.com/2d0j4wg.png a pullstring walker is in the middle of a string that is connected to two mountain sides. the string formes an angle of 10 degrees under the horizontal. the walker has a mass of 50 KG. find the tension of the string any...

so? is it correct or not!

yes. 21KG(10M/S^2) -FT= ma 210-FT=21(2.75) FT=152.25

how about the tension on the cord between A and B ft=152.25 ?

i thought they're supposed to show the work?

can anybody check my answer?

a guy already answered the question on yahoo answers!

210-ft=57.75 ft=152.25 ?

ok I showed my work for that part. can anyone tell me if 367.5 is correct?

correct my bad got confused. using different weights in one problem does not make sense either. so 210-ft= 2.75(210) I got ft= 367.5 ?

well since we don't have an angle. then 210-ft= 2.75(5) so 210-ft= 13.75 -ft = 13.75-210 ft= 196.25 ? is that correct!

I figured out the acceleration part but I can't figure out the Tension!