Question

Find the expression for acceleration in a string when two bodies are hanging vertically attached

with a string passing over a frictionless pulley.​

Answer 1

$\huge \mathbb{ \red{★᭄ꦿ᭄ANS}\pink{W}\purple{E}\blue{R}\green{★᭄ꦿ᭄}}$

☛☞For mass m, the forces are given as

For mass m, the forces are given asT−mg=ma (1)

For mass m, the forces are given asT−mg=ma (1) Similarly, for mass M, the forces are given as

For mass m, the forces are given asT−mg=ma (1) Similarly, for mass M, the forces are given asMg−T=Ma From equation (1), we have

For mass m, the forces are given asT−mg=ma (1) Similarly, for mass M, the forces are given asMg−T=Ma From equation (1), we haveT=mg+ma (3)

For mass m, the forces are given asT−mg=ma (1) Similarly, for mass M, the forces are given asMg−T=Ma From equation (1), we haveT=mg+ma (3) Substituting equation (3) in (2) we get

For mass m, the forces are given asT−mg=ma (1) Similarly, for mass M, the forces are given asMg−T=Ma From equation (1), we haveT=mg+ma (3) Substituting equation (3) in (2) we getMg−mg−ma=Mag(M−m)=a(M+m)a=( M+mM−m)g

For mass m, the forces are given asT−mg=ma (1) Similarly, for mass M, the forces are given asMg−T=Ma From equation (1), we haveT=mg+ma (3) Substituting equation (3) in (2) we getMg−mg−ma=Mag(M−m)=a(M+m)a=( M+mM−m)g T=m×( M+m

For mass m, the forces are given asT−mg=ma (1) Similarly, for mass M, the forces are given asMg−T=Ma From equation (1), we haveT=mg+ma (3) Substituting equation (3) in (2) we getMg−mg−ma=Mag(M−m)=a(M+m)a=( M+mM−m)g T=m×( M+mM−m)g+mg

)g+mg T= M+mMm−m 2 g+mg

)g+mg T= M+mMm−m 2 g+mg T=g( M+mMm−m 2+m)

)g+mg T= M+mMm−m 2 g+mg T=g( M+mMm−m 2+m)T=g( m+mMm−m2+m 2+Mm)=g( M+m2Mm)