6.The minimum value of coefficient of friction `(mu)` such that block of mass '5kg' remains at rest is [NCERT Pg.102]

A block of mass M is placed on a rough horizontal surface. It is connected by means of a light inextensible string passing over a smooth pulley . The other end of string is connected to a block of mass m. There is no friction between vertical surface and block m The minimum value of coefficient of friction mu such that system remains at rest , is

Two identical blocks 'A' and 'B' each of mass 5kg are connected with an ideal string and placed on a rough horizontal table having coefficient of friction (mu = 0.4 ) third block of mass 'm' are connected to block 'B' with another ideal string which is passing over a ideal pully as shown in figure.The minimum value of mass of block 'C' such that blocks just starts moving is (g= 10 m/ s^2 )

Find the minimum value of horizontal force (F) required on the block of mass m to keep it at rest on the wall. Given the coefficient of friction between the surfaces is mu .

In the figure shown, the coefficient of static friction between the block A of mass 20 kg and horizontal table is 0.2. What should be the minimum mass of hanging block just beyond which blocks start moving?

In the diagram shown, the block A and B are of the same mass M and the mass of the block C is (M_(1)) . Friction is present only under the block A. the whole system is suddenly released from the state of rest. The minimum coefficient of friction on block A to keep it in the state of rest is equal to .

A horizontal force F is applied at the top of an equilateral triangular block having mass m. The minimum coefficient of friction required to topple the block before translation is mu ,find 100 mu ? ( sqrt(3) = 1.74)

In the adjoining figure, the coefficient of friction between wedge (of mass M ) and block (of mass m ) is mu . Find the minimum horizontal force F required to keep the block stationary with respect to wedge.

Block A of mass 35kg is resting on a frictionless floor. Another block B of mass 7kg is resting on it as shown in the figure. The coefficient of friction between the blocks is 0.5 while kinetic friction is 0.4. (g = 10ms^(-2)) The minimum value of force to cause block B to move with respect to block Ais :

The minimum value of mass m for which the 100 kg block can remain in static equilibrium is

Block B of mass 2 kg rests on block A of mass 10 kg. All surfaces are rough with the value of coefficient of friction as shown in the figure. Find the minimum force F that should ber appplied on blocks A to cause relative motion between A and B. (g=10(m)/(s^2) )