In the arrangement shown in show that tension in the string between masses `m_(2)` and `m_(3)` is `T = (2 m_(1) m_(3) g)/(m_(1) + m_(2) + m_(3))`

In the arrangement as shown, tension T_(2) is (g=10m//s^(2))

In the arrangement as shown, tension T_(2) is - (g=10m//s^(2))

In the pulley system shown here, find the relationship between acceleration of m_(1),m_(2) and m_(3)

In the arrangement shown in the all contact surfaces are smooth strings and pulleys are massless Given M_(1) =1kg, M_(2) =2kg, M_(3) =4kg and g =10ms^(-2) .

The arrangement shown in the figure, the systme of masses m_(1), m_(2) and m_(3) is being pushed by a force F applied on m_(1) horizontally. In order to prevent the downwards slipping of m_(2) between m_(1) and m_(3) if coefficient of friction between m_(2) and m_(3) is mu and all the other surface are smooth, the minimum value of F :

Consider the system ofpulleysasshown in figure-2.197. Find the acceleration of the three masses m^(1),m^(2) and m^(3). (m_(1) = I kg,m_(2) = 2kg " and " m_(3) =3kg)

Two cubes of masses m_(1) and m_(2) lie on two frictionless slopes of the block A which rests on horizontal table. The cubes are connected by a string, which passes over a pulley as shown in the figure-2.67. To what horizontal acceleration "f" the block accelerates so that the cubes do not slide down the planes ? What is the tension in the string in this situation ? [f=g((m_(1)sinalpha+m_(2)sinbeta)/(m_(1)cosalpha+m_(2)cosbeta)),T = (m_(1)m_(2)gsin(alpha-beta))/(m_(1)cosalpha+m_(2)cosbeta)]

In the figure shown tension in string AB always lies between m_(1)g and m_(2)g . (m_(1)!=m_(2)) Tension in massless string is uniform throughout.