In the arrangement shown- neglect the mass of the ropes and pulley- What must be the value of M - m to keep the system in equilibrium-Assume that there is no friction anywhere and tension due to string on M is along the plane-

As the block m is in equilibrium nbsp; ⟹mg=T”c For the bottommost pulley, we can say T'=2T” nbsp; and 2T=T' ( as pulleys are assumed to be light) nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Tension in a String

In the arrang...

Question

In the arrangement shown, neglect the mass of the ropes and pulley. What must be the value of $\frac{M}{m}$ to keep the system in equilibrium?
(Assume that there is no friction anywhere and tension due to string on $M$ is along the plane).

Open in App

Solution

As the block $m$ is in equilibrium
$⟹ m g = T^{''} c$
For the bottommost pulley, we can say $T^{'} = 2 T^{''}$ and $2 T = T^{'}$
( as pulleys are assumed to be light)
$∴ T = T^{''} = m g$
For mass $M$ on the inclined plane, we can say :
$T = M g sin 30 = M g / 2$
$m g = M g / 2$
$∴ m = M / 2$

Suggest Corrections

Similar questions

Q. In the arrangement shown, neglect the mass of the ropes and pulley. What must be the value of

\frac{M}{m}

to keep the system in equilibrium?
(Assume that there is no friction anywhere and tension due to string on

M

is along the plane).

Q. In the arrangement shown, neglect the mass of the ropes and pulley. What must be the value of

m

to keep the system in equilibrium? There is no friction anywhere.

Q. In the arrangement shown, neglect the mass of the ropes and pulley. What must be the value of

m

to keep the system in equilibrium. There is no friction anywhere.
Given: For

M

, force acting down the plane is

M g sin 30^{\circ}

Q. In the arrangement shown, neglect the mass of the ropes and the pulley. What must be the value of

m

to keep the system in equilibrium? (Neglect friction every where).

Q. In the arrangement shown in figure,find the tensions in the rope and accelerations of the masses

m_{1}

and

m_{2}

and pulleys

p_{1}

and

p_{2}

when the system is set free to move. Assume the pulleys to be massless and strings are light and inextensible.

Join BYJU'S Learning Program

In the arrangement shown, neglect the mass of the ropes and pulley. What must be the value of Mm to keep the system in equilibrium? (Assume that there is no friction anywhere and tension due to string on M is along the plane).

As the block m is in equilibrium ⟹mg=T′′c For the bottommost pulley, we can say T′=2T′′ and 2T=T′ ( as pulleys are assumed to be light) ∴T=T′′=mg For mass M on the inclined plane, we can say : T=Mgsin30=Mg/2 mg=Mg/2 ∴m=M/2

In the arrangement shown, neglect the mass of the ropes and pulley. What must be the value of $\frac{M}{m}$ to keep the system in equilibrium?
(Assume that there is no friction anywhere and tension due to string on $M$ is along the plane).

As the block $m$ is in equilibrium
$⟹ m g = T^{''} c$
For the bottommost pulley, we can say $T^{'} = 2 T^{''}$ and $2 T = T^{'}$
( as pulleys are assumed to be light)
$∴ T = T^{''} = m g$
For mass $M$ on the inclined plane, we can say :
$T = M g sin 30 = M g / 2$
$m g = M g / 2$
$∴ m = M / 2$