# Homework Help: Inclined slope, b degrees

1. Nov 26, 2008

### devanlevin

as in the diagram below, mass m1 is placed above mass m2 on a sloped incline with an angle of b degrees. the frictional coefficient between the two masses is C1 and the coeficient between mass 2 and the slope is C2.
find the acceleration of the masses.

after dividing the vectors into components i said
N1=(m1)gcosb
f1=(C1)(m1)gcosb

N2=N1+(M2)gcosb=(M1+M2)gcosb
f2=(C2)(M1+M2)gcosb

for mass 1
F=m1gsinb-f1=m1a
=(m1)gsinb-(C1)(m1)gcosb=m1a

a1=g(sinb-C1cosb)=gcosb(tgb-C1)
meanin that when C1 is equal to tanb, body 1 will not move relative to body 2 or will not move relative to the slope?

now for body2

F=m2gsinb+f1-f2=m2a------> is this correct??
after opening i get

a2=m2gcosb(tgb-C2)+m1gcosb(C1-C2) is this correct?

Last edited by a moderator: May 3, 2017
2. Nov 26, 2008

### Redbelly98

Staff Emeritus
The expression for a1 is correct. However, when C1=tan(b), a1 will be zero. a1 is not the velocity, it is the ______ .

Should be "m2 a2" on the left side. Otherwise it looks good.

3. Nov 27, 2008

### devanlevin

a1 is the acceleration, sorry, m1 will not accelerate, but is it relative to the slope or rellative to m2, ie will a1 relative to the a person standing on the slope be
1) a1=g(sinb-C1cosb)=gcosb(tgb-C1) or
2) a1=g(sinb-C1cosb)=gcosb(tgb-C1) + a2

and a2=[m2gcosb(tgb-C2)+m1gcosb(C1-C2)]m2

so a2=gcosb[(tgb-C2)+(m1/m2)(c1-C2)] relative to the slope

4. Nov 27, 2008

### Redbelly98

Staff Emeritus
a1: Since you have accounted for all the forces when you got a=Fnet/m, the expression you had is the acceleration. (1) is correct.

a2: looks good.