1) Tension in string between P and Q = T1
tension in string between Q and R = T2
let all of them move with a uniform acceleration of a.
let mass of P , Q and R be m.
F - T1 = m a for P
T1 - T2 = m a for Q
T2 = m a for R
=> T1 = ma + ma = 2 ma
F = 3 ma
ratio = 2
T1 = m g
T1 Sin Ф + T2 Cos Ф = √2 m g
T2 = m g
Sin Ф + Cos Ф = √2
square it, 1 + sin 2Ф = 2
Sin 2Ф = 1 => Ф = 45 deg
please see diagram for free body diagram of pulley.
there are vertical forces M g (same as tension T) and m g weight of pulley. there is a horizontal force Mg = Tension.
resultant force = √ [ (Mg+mg)² + (Mg)² ]
same as your expected answer
4) inertia of motion. There is not enough centripetal force acting on them. So they continue to move in a straight line.
Center of mass of body m and plank of mass M is at [ M * L/2 + m * 0 ] / (M+m) , from the end where the body m is present.
that is = CM at ML/2(M+m)
We assume there is no friction between water and plank.
When the body goes to the other end, the center of mass will be at ML/2(M+m) from that end.
Since there is no external force on the two body system together, the CM stays at the same point.
Hence the plank moves by L - ML/2(M+m) - ML/2(M+m) = L * m /(M+m)