- #1
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.
http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273088118963403586
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?
find the acceleration of the masses.
http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273088118963403586
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: