:approve: okay
i drew it that way 'cause that's what my prof does. hmmm, bad habit i guess. tsk tsk.
so since F32 and F23 have different signs, that completely makes sense then! thank you really!
hey cool!
erm...those negative signs came from the way i drew my free body diagram. i put space in between each crate. so concentrating on the 3rd crate, i have horizontal forces of f_k to the left and another leftward force of crate 3 acting on crate 2 (F23). this force of which should be...
hmm...worked on it some more, and I'm getting:
F32 = -m3*a_x-f_k with f_k=(mu)_k * m3*g
= -(20kg*0.473m/s^2)-(0.700*20kg*9.8m/s^2)
= -147 N
therefore, magnitude is 147 N
-possibly :eek:
okies, i did not see that the question was actually asking for magnitude. woops.
let's focus on crate 3. so i'd have to do the x component/y component situation. but, wouldn't the y component be zero? normal force minus m3*g?
so then for the x component:
F_x = F32-F23-f_k
with F23 being...
Homework Statement
The figure shows 3 crates being pushed over a concrete floor by a horizontal force f of magnitude 440N. The masses of the crates are m1=30 kg, m2=10kg, and m3=20kg. The coefficient of kinetic friction between the floor and each of the crates is 0.7. what is the magnitude F32...
Thank you Mindscrape.
Yes the tension equation is from Newton's second law. I had to research the Galilean relativity...ermmm. I'll just ask my professor later. Thank you though!
Homework Statement
In chapter 5, problems 26:
A provided figure shows an overhead view of a 0.0250 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F_1 has a magnitude of 6.00 N and is at θ_1 = 30.0° (and, btw, is in quadrant II)...
Hello.
Homework Statement
There is a link to the problem and its picture here: http://media.wiley.com/product_data/excerpt/19/04717580/0471758019-1.pdf
It is # 46.
Here is the problem as well:
In Fig. 4-44, a ball is thrown up onto a roof, landing 4.00 s later at height h=20.0m...