TS EAMCET 5-Aug-2021 Shift 2 Question Paper

The given situation is shown below. Coefficient of friction, µ=0.5, force applied by person on wall, N=500N Let, mass of the person be m.
Here, f1,f2 and w be the friction forces and weight of man respectively.
As the person have to be at rest i.e. equilibrium
Hence, ‌‌f1+f2=mg and ‌‌f=µN
∴‌‌µN1+µN2=mg
Since, ‌‌N1=N2=N=500N
∴‌‌2µN=mg
⇒2×‌

5

10

×500=m×10⇒m=50kg