hey! okay! soo nowwww, i would thinkk,
Ff1 = (7.47)(9.8)(0.172) = 12.6 N
Ff2 = (11.5 + 7.47)(9.8)(0.172) = 32 N
80.3 - 12.6 - 32 = 11.5 a
a = 3.12 m/s^2 :( this is what i had earlier :(
okk, i see what u mean, there's friction at the top and bottom surfaces, but which direction does each friction force move in? wouldn't they BOTH move opposite to the horizontal force? so would the x direction equation be:
F - Ff1 - Ff2 = m2a
or am i completely missing something...because...
im confused why the friction is top and bottom...and am i making equations of the x and y components? because if that is the case, then there is only 1 force in the x direction (F)
i don't get it :(
:( hmm, okk
horizontal force = right
force of gravity = bottom
force of friction = left
force normal = top
tension = left
im missing a force :O i can't figure out which one that iss :(
so the FBD for the lower block, would be horizontal arrow forward, friction backward, weight of both masses downward, normal upward AND tension backward? I am so unsure about the tension...
so this is what i gather,
Fn = (m1 + m2)(9.8) = 186 N??
Ff = 186 x 0.172 = 32 N on lower block...
Oh! So I calculated the frictional force on the bottom block (19.38 N), but I am still confused how to incorporate it into the equation. I am getting thrown off by the fact that the tension is only on the top block and i don't know which forces to add and subtract.
Could it be:
80.3 - 19.4 -...
Okay, I thought it might be:
Fnet = ma
Fa - Ff = m2a
(80.3) - (12.6) = (11.5)a
a=5.89 m/s^2
But this is wrong! Am i missing a negative? or a whole force? so frustrating! lol!
Homework Statement
A m1=7.47 kg block is placed on top of a m2=11.5 kg block. A horizontal force of 80.3 N is applied to the 11.5 kg block causing it to slide. The 7.47 kg block is tied to the wall by a string. The co-efficient of kinetic friction between all surfaces is 0.172.
a)...