The answer to this problem should act as an eye opener to the simplicity of many things;
Assume: m1 moves downwards. Find the accn. of m1.
Bet you had to frame some (2) equations...i don't...
Try to observe something in the result. Report anything unusual;
hmm..friction on both sides,m2 are opposing the accl......m1 is supporting...so we get this kind of result
ReplyDeletecud not observe anything more
a= g(m1(sinA-u1cosA)-m2(sinB+u2cosB))/m1+m2
that was all i cud observe too..
ReplyDeletetake forces along the thread on the system:
ReplyDeleteF=(m1gSinA)-(u1m1gCosA)-(u2m2gCosA)-(m2gSinA)
M=m1+m2;
divide F by M to write the ans..
yes bhaiya it is simple. ACCELERATION OF THE SYSTEM= (NET FORCE ALONG LINE OF MOTION)/(TOTAL MASS IN MOTION). HERE LINE OF MOTION IS ALONG THREAD
ReplyDelete