Box 1:
m = 1.55
mg = 1.55*9.81 = 15.2 N
friction force up slope = .226*15.2 cos 30 = 2.98 N up slope
weight component down slope = 15.2 sin 30
= 7.6 N down slope
T down slope (deal with after we find a
Box2:
m = 3.1
mg = 3.1*9.81 = 30.4 N
friction force up slope = .113*30.4 cos 30 = 2.98 N up slope (same as other one)
weight force down slope = 30.4 sin 30 = 15.2 N
T is up slope
so
do whole system together as rigid body to find a
ma = (1.55+3.1)a = 7.6+15.2 - 2.98 - 2.98
a = 3.62 m/s^2
Now do either mass to get T
Box 1
7.6 + T - 2.98 = 1.55 (3.62)
T = 1.21 N
two blocks are moving down a ramp that is inclined at 30.0. Box 1(of mass 1.55kg) is above box 2(of 3.1kg) and they are connected by a massless rod. Find the tension of the rod if the coefficient of kinetic friction for box 1 is .226 and box 2 is .113
my work so far:
a=m2gsintheta-migsintheta-mk2m2gcostheta-mk1m1gcostheta/mt
=0.355m/s^2 (I'm not sure if this is the right answer)
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