Problem on an inclined plane

1. Feb 16, 2010

haengbon

1. The problem statement, all variables and given/known data

A 3kg load on an inclined plane of angel 30 degrees is connected by a cord over a frictionless pulley to a second load which has a mass of 2 kg. Find the:

b) tension of each chord

2. Relevant equations

none?

3. The attempt at a solution

block A / box on inclined plane

$$\sum$$Fy= 0
-Wycos30$$\circ$$ + NF= 0
-30cos30 + NF = 0
-25.98 + NF = 0
-25.98 = NF
NF = 25.98

f=$$\mu$$ NF
f= 0.3 (25.98)
f= 7.794

$$\sum$$Fx= ma
Wxsin30$$\circ$$ - T - f = ma
15 - T - 7.794 = ma
15 - 7.794 - T = (3)(a)
-T = 3a - 7.206
T = 7.206 - 3a

block b

$$\sum$$Fx= ma
T-W = ma
T - (2)(10) = (2)a
T - 20 = 2a
T= 2a+20

T=T
2a+20 = 7.206 - 3a
2a+3a = 7.206 - 20
5a = -12.794
a = - 2.56 m/s2

T1=7.206 - 3a
T= 7.206- 3(-2.56)
T= 14.886 N

T2= 2a+20
T= 2(-2.56) + 20
T= -3.1176 N

T2+T1 = TotalT
14.886 + (-3.1176) = TotalT
TotalT= 11.7684 N

I'm not sure if I got the right answer :( someone help?

2. Feb 16, 2010

rl.bhat

Loads are connected by a single cord. So T1 = T2 = T.

3. Feb 16, 2010

haengbon

so the tensions I got are incorrect then?
may I ask if the acceleration is correct?

4. Feb 16, 2010