TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

The given situation is shown in the figure.Resolving applied force, we have following forces on block
Given, tan‌θ=‌

3

4

,sin‌θ=‌

3

5

and cos‌θ=‌

4

5

So, horizontal component,
F1=12×cos‌θ=12×‌

4

5

=‌

48

5

N
and vertical component,
F2=12×sin‌θ=12×‌

3

5

=‌

36

5

N
So, total applied force on ground is
10+‌

36

5

+2×10=37.2N
So, reaction N1 of ground =37.2N upwards
Also, total force horizontally on wall is
F1=‌

48

5

=9.6N
So, reaction N2 of wall =9.6N towards left