An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the prism as shown in the figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is :
Toppling will take place about right most point so,
F ×(√3a/2) = mg × a/2
⇒ F= mg/√3
A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its centre
As the object slides with constant speed friction force is balancing mgsinθ. Torque of Normal reaction and friction force about centre of object will cancel each other.
τ=mg.sinθ × a/2 = (mg a sinθ)/2
A wheel having moment of inertia 2 kg-m² about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in one minute would be
ω = ω0 + α.t
0= (60 × 2π /60) + α.60
⇒ α = - π/30
Torque = I α = (2× π/30) = (π/15) N-m
The figure shows a horizontal block of mass M suspended by two wires A and B. The centre of mass of the block is closer to B than A. (i) Is the magnitude of the torque due to wire A is greater, less or equal to that due to B w.r.t. centre of mass ? (ii) Which wire A or B exerts more force on the block ?
As the object is in rotational equilibrium, Net torque acting on the object is zero.
so, Torque of TA = Torque of TB
TAXA= TBXB
\[ \frac{T_{A}}{T_{B}}=\frac{X_{B}}{X_{A}} \]
\[ X_{B} < X_{A} \]
\[ T_{A} < T_{B} \]