A block of mass M is pulled on a smooth horizontal table by a string making an angle θ with the horizontal as shown in figure. If the acceleration of the block is 'a', find the force applied by the string and by the table on the block.
Let us consider the block as the system.
The forces on the block are
(a) Pull of the earth, Mg, vertically downward,
(b) Contact force by the table, N, vertically upward,
(c) Pull of the string, T, along the string.
Let's draw blocks'Free Body Diagram
The acceleration of the block is horizontal and towards the right. Take this direction as the +vex-axis and vertically upward direction as +ve Y-axis. We have,
Component of Mg along the X-axis = 0
Component of N along the X-axis = 0
Component of T along the X-axis = T cos θ
Hence the total force along the X-axis = T cos θ.
Using Newton's II law, T cos θ = Ma. . . . . . . (i)
Component of Mg along the Y-axis = - Mg
Component of N along the Y-axis = N
Component of T along the Y-axis = T sin θ.
Total force along the Y-axis(as there is no net acceleration along this direction) = N + T sin θ - Mg
Again Using Newton's II law along assumed Y axis, N + T sin θ = Mg - 0 . . . . . . (ii)
From equation (i), T=Macosθ.Putting this in equation(ii)
N=Mg−Ma tanθ.