Question

Block A in the figure below has mass mA = 5.0 kg and is sliding down the ramp. Block B has mass mB = 1.8 kg. The coefficient of kinetic friction between block B and the horizontal plane is 0.50. The inclined plane is frictionless and at angle 30°.

Answer 1

find what?

Answer 2

Equations i have:
(x1)Ft-Fk=m1a
FT=m1a+(.18*29.4N)
FT=(3*3.08) + 5.29=14.5
(x2) m2gsin(30)-FT=m2a
(x2) 39.2-m1a+(.18*29.4N) =m2a
(x2) 39.2-5.29=a(m1+m2) =11a
33.9/11=a=3.08
(y1)FN-m1g=0
FN=m1g=29.4N

for some reason whenever i come here and write out my work i always see my mistake. maybe i should just type out every problem and save my self the time writing it out 3 times.

Answer 3

lol i even paseted the wrong problem

Answer 4

(mB) : (ox) : T - f =mB*a
\[f =\mu* mB*g\]
we get T - 0.5*mB*g = mB*a (1)
(mA) :(ox) \[-T + mA*g \sin \theta =mA*a (2)\]
take (1) + (2)
a(mA +mB) = -0.5*mB*g +mA*g*sin30
a = 2.3m/s^2