In the arrangement, shown in figure, pulleys A and B are massless and frictionless and threads are ideal. Block of mass m1 will remain at rest if:
In the movable pulley system, tension in the string connecting m2 and m3 is:
Since this tension acts twice to balance
, we equate:
Cancelling
and rearranging gives:
A 1 kg object strikes a wall with velocity \(1\text{ ms}^{-1}\) at an angle of \(60^\circ\) with the wall and reflects at the same angle. If it remains in contact with wall for 0.1 s, then the force exerted on the wall is
The angle with the normal is \(90^\circ - 60^\circ = 30^\circ\). The change in momentum perpendicular to the wall is \(\Delta p = 2mv\cos(30^\circ) = 2(1)(1)\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}\text{ kg m s}^{-1}\). Average force \(F = \frac{\Delta p}{\Delta t} = \frac{\sqrt{3}}{0.1} = 10\sqrt{3}\text{ N}\).
A \(3.0\text{ kg}\) mass is moving in a plane, with its x and y coordinates given by \(x = 24t^2 – 1\) and \(y = 3t^3 + 2\), where x and y are in meters and t is in second. Find the magnitude of the net force acting on this mass at \(t = 2\text{ sec}\).
Differentiating position equations twice yields accelerations: \(a_x = 48\text{ m/s}^2\) and \(a_y = 18t\text{ m/s}^2\). At \(t = 2\text{ s}\), \(a_y = 36\text{ m/s}^2\). Total acceleration \(a = \sqrt{a_x^2 + a_y^2} = 60\text{ m/s}^2\). The net force is \(F = ma = 3.0 \times 60 = 180\text{ N}\).
A particle moves on a rough horizontal ground with some initial velocity say \(v_0\). If \(3/4^{\text{th}}\) of its kinetic energy is lost in friction in time \(t_0\). Then coefficient of friction between the particle and the ground is:
Since \(3/4^{\text{th}}\) of kinetic energy is lost, the remaining kinetic energy is \(1/4^{\text{th}}\), meaning the final velocity \(v = v_0/2\). Using \(v = v_0 - at_0\) where \(a = \mu g\), we get \(v_0/2 = v_0 - \mu gt_0 ⇒ \mu = \frac{v_0}{2gt_0}\).
A body of mass 1 kg has velocity \(1\text{ ms}^{-1}\), up an inclined plane of angle of \(30^\circ\) to the horizontal. The friction coefficient is \(\frac{1}{\sqrt{3}}\). The distance the body travels before stopping is (\(g = 10\text{ m s}^{-2}\)):
Retardation \(a = g(\sin\theta + \mu\cos\theta) = 10(\sin 30^\circ + \frac{1}{\sqrt{3}}\cos 30^\circ) = 10(0.5 + 0.5) = 10\text{ m/s}^2\). Using \(v^2 = u^2 - 2as ⇒ 0 = 1^2 - 2(10)s\), we get \(s = 0.05\text{ m} = 5\text{ cm}\).
A uniform chain of length \( L \) is placed on a rough horizontal table with some part hanging below the table. If the length of the hanging part becomes \( \frac{2L}{5} \) then the chain starts sliding on the table.
The co-efficient of friction between the chain and the table is:
The hanging part of length \( \frac{2L}{5} \) exerts a pulling force of \( \frac{2}{5} Mg \). The part on the table of length \( \frac{3L}{5} \) experiences a maximum friction force of \( \mu \frac{3}{5} Mg \). Equating the forces at the verge of sliding gives \( \mu = \frac{2}{3} \).