Fig A.43 show three block on a rough surface under the influence of a force $P$ of the same magnitude in all the three cases.
Co0efficient of friction is the same between each block and the ground. What possible relation holds between the magnitude of normal
reaction and friction forces? (Assume that the block do not overturn about edge.) Here, $f_{A}$ , $f_{B}$ and $f_{C}$ are frictional forces and $N_{A}$ , $N_{B}$ and $N_{C}$ are reactions.

N_{A} > N_{C} > N_{B}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f_{A} > f_{C} > f_{B}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f_{C} > f_{A} = f_{B}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

N_{C} > N_{A} = N_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $N_{A} > N_{C} > N_{B}$
B $f_{A} > f_{C} > f_{B}$
C $f_{C} > f_{A} = f_{B}$
$N_{A} = m g + P sin θ$
$N_{B} = m g - P sin θ$
$N_{C} = m g$
Clearly, $N_{A} > N_{C} > N_{B}$
Now to find relation between frictions:
(i) Let sliding does not occur:
$f_{A} = P cos θ, f_{b} = P cos θ, f_{C} = P$
$\Rightarrow f_{C} > f_{A} = f_{B}$
(ii) Let sliding occur at all contact surfaces : $f_{A} = μ N_{A}, f_{B} = μ N_{B}, f_{C} = μ N_{C}$
$\Rightarrow f_{A} > f_{C} > f_{B}$

Suggest Corrections

Similar questions

Three identical vessels A, B and C contain same quantity of liquid. In each vessel balls of different densities but same masses are placed. In vessel A, the ball is partly submerged; in vessel B, the ball is completely submerged but floating and in vessel C, the ball has sunk to the base. If $F_{A}$ , $F_{B}$ and $F_{C}$ are the total forces acting on the base of vessels A, B and C respectively, then