FRICTION !!!!!!!! HELP!!!!

[nid404]

In fig m1=4m2. Pulleys are smooth and the thread is light. At t=0 system is released from rest. Find the maximum height reached by m2

[astarmathsandphysics]

Will answer when i get home

[nid404]

Will answer when i get home

k thank you

[astarmathsandphysics]

In fig m1=4m2. Pulleys are smooth and the thread is light. At t=0 system is released from rest. Find the maximum height reached by m2

since m1 has two strings attached it has half the acc
4mg-2T=2ma  (1) for m1
T-mg=ma    (2) for m2
(1)+2*(2)
2mg=4ma a=g/2
fpor m1 a=g/4=2.45 v^2=u^2+2as v2=0+2*9.8/4*0.2=0.98 v=sqrt(0.98)=0.99
The speed of m2 is twice the speed of m1 ie 1.98
after m1 stops m2 takes on acceleration -g v^2=u^2+2as s=(0-3.96)/2*9.8=0.2
height reached=0.4+0.2=0.6

[nid404]

In fig m1=4m2. Pulleys are smooth and the thread is light. At t=0 system is released from rest. Find the maximum height reached by m2

since m1 has two strings attached it has half the acc
4mg-2T=2ma  (1) for m1
T-mg=ma    (2) for m2
(1)+2*(2)
2mg=4ma a=g/2
fpor m1 v^2=u^2+2as v2=0+2*9.8/2*0.2=1.96 v=sqrt(1.96)=1.4
after m1 stops m2 takes on acceleration -g v^2=u^2+2as s=(0-1.96)/2*9.8=0.1
height reached=0.2+0.1=0.3

answer is 6.6 cm

[astarmathsandphysics]

I looked at my post again. Did you mean 0.6m?

[nid404]

I looked at my post again. Did you mean 0.6m?

yup sorry...0.6 cm

[astarmathsandphysics]

I corrected my post.

[nid404]

thank you sir

I have some more comin

I'm bad at this.....i hate constraints


Please help me in this one too
In the arrangement shown, the mass of the bar and wedge is same and wedge angle is 37 degrees. The  masses of the pulley and thread r negligible. Friction is absent. Find the acceleration of the wedge

[astarmathsandphysics]

Will answer when i get home

[nid404]

Will answer when i get home

cool

[astarmathsandphysics]

Resolving horizontally for the wedge

T+mgsin37-Tcos37=ma ma=mgsin37+T(1-cos37) (1)
resolving down the slope for the block
mgsin37-T=ma/cos37 so T=mgsin37-ma/cos37 sub into (1)
ma=mgsin37+(mgsin37-ma/cos37)(1-cos37)
then divide by m

 

[nid404]

Resolving horizontally for the wedge

T+mgsin37-Tcos37=ma ma=mgsin37+T(1-cos37) (1)
resolving down the slope for the block
mgsin37-T=ma/cos37 so T=mgsin37-ma/cos37 sub into (1)
ma=mgsin37+(mgsin37-ma/cos37)(1-cos37)
then divide by m

 

I know this may kinda irritate you......but when I solve this...I get 5g/13... ........the answer is 13mg/18..damn!!!

[nid404]

could u draw a free body diagram please....I won't understand this otherwise

[astarmathsandphysics]

Resolving horizontally for the wedge

T+mgsin37-Tcos37=ma ma=mgsin37+T(1-cos37) (1)
resolving down the slope for the block
mgsin37-T=ma so T=mgsin37-ma sub into (1)
ma=mgsin37+(mgsin37-ma)(1-cos37)
ma+ma(1-cos37)=mgsin37+mg(1-cos37) so a=(mgsin37+mg(1-cos37) )/(m+m(1-cos37))