What is the tension and mass relationship in a three-block system on an incline?

[oreosama]

Homework Statement

blocks A B AND C are connected as show. block A and B have the same mass "m" and the coefficient of kinetic friction between each block is uk. block C descends with constant velocity. use 30 degrees for the angle of the incline.

given m, uk
find

a.
tension in rope between blocks A AND B

b.

mass of block C

c.

if the rope connected to block A is cut, determine the accel of block C

The Attempt at a Solution

http://i.imgur.com/KJsro.png

FAx = T1 - uk*N1 = 0
FAy = N1 - mg = 0

FBx = T2 - T1 - uk*N2 = 0
FBy = N - mg*cos(30) = 0

FCy = mc*g - T2 = 0


a.

N1 = mg -> T1 = uk*mg


b.

N2 = mg*cos(30) - > T2 = uk*mg + uk*mg*cos(30)

mc*g - ( uk*mg + uk*mg*cos(30) ) = 0

mc = uk*m + uk*m*sqrt(3)/2

I think these are right(confirm pls)


c is what confuses me but my attempt:

FBx = T2 - uk*N2 - g*sin(30) = 0
FBy = N2 = mg*cos(30)

FCy = mc*g - T2 = mc*a


T2 = uk*mg*cos(30) + g*sin(30)

a = (mc * g - T2) / mc * g


(uk*m + uk*m*sqrt(3)/2 - uk*m*sqrt(3)/2 + 1/2 ) / uk*m + uk*m*sqrt(3)/2 (g should cancel out)

is this all right? thanks for any help.

 

[azizlwl]

a. Correct t1=mgμ_k
b. t2 is balanced by 3 forces, t1, friction and the weight of the mass.
c. t3 is balance by weight of mass c.

 

[PhanthomJay]

c is what confuses me but my attempt:

FBx = T2 - uk*N2 - m[/color]g*sin(30) = 0

Note omission in red. Also, block B is accelerating, correct? Why have you set the sum of forces equal to 0??

FBy = N2 = mg*cos(30)

yes

FCy = mc*g - T2 = mc*a

yes

T2 = uk*mg*cos(30) + g*sin(30)

please correct

a = (mc * g - T2) / mc * g

where did the g come from?

(uk*m + uk*m*sqrt(3)/2 - uk*m*sqrt(3)/2 + 1/2 ) / uk*m + uk*m*sqrt(3)/2 (g should cancel out)

is this all right? thanks for any help.

Make appropriate corrections.

 

[oreosama]

T1 = uk*mg

T2 - T1 - Ff - mgsin(30) = 0

T2 = uk*mg + uk*mg*cos(30) + mg*sin(30)

mcg - T2 = 0

mc = uk*m + uk*m*sqrt(3)/2 + m/2c)

FBx = T2 - uk*mg*cos(30) - mg*sin(30) = ma

FCy = mc*g - T2 = mc*a

T2 = mc*a + mc*gmc*a + mc*g - uk*mg*cos(30) - mg*sin(30) = ma

mc*g - uk*mg*cos(30) - mg*sin(30) = ma - mc*a

mc*g - uk*mg*cos(30) - mg*sin(30) = a(m - mc)a = (mc*g - uk*mg*sqrt(3)/2 - mg/2) / (m - mc)

a = (uk*mg + uk*mg*sqrt(3)/2 + mg/2 - uk*mg*sqrt(3)/2 - mg/2) /( m - mc)a = (uk*mg) /( m - uk*m - uk*m*sqrt(3)/2 - m/2)

 

[PhanthomJay]

T1 = uk*mg

T2 - T1 - Ff - mgsin(30) = 0

T2 = uk*mg + uk*mg*cos(30) + mg*sin(30)

mcg - T2 = 0

mc = uk*m + uk*m*sqrt(3)/2 + m/2


c)

FBx = T2 - uk*mg*cos(30) - mg*sin(30) = ma

FCy = mc*g - T2 = mc*a

OK

T2 = mc*a + mc*g

You have a signage error here. Check algebra.

mc*a + mc*g - uk*mg*cos(30) - mg*sin(30) = ma

mc*g - uk*mg*cos(30) - mg*sin(30) = ma - mc*a

mc*g - uk*mg*cos(30) - mg*sin(30) = a(m - mc)


a = (mc*g - uk*mg*sqrt(3)/2 - mg/2) / (m - mc)

this should be OK once you make the correction on your algebra/signage error noted above when determining T2.

a = (uk*mg + uk*mg*sqrt(3)/2 + mg/2 - uk*mg*sqrt(3)/2 - mg/2) /( m - mc)


a = (uk*mg) /( m - uk*m - uk*m*sqrt(3)/2 - m/2)

These last 2 steps are OK where you make the substitution for mc