Question

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative to Mc. Ignore all friction. Assume Mb does not make contact with Mc.

Answer 1

The magnitude of the force F is given by

F = (
`M_(a)` +
`M_(b)` +
`M_(c)` ) *(
`M_(b)`*g/(
`\sqrt{M_(a) ^(2)-M_(b) ^(2)}`))

Step-by-step explanation:

Given there are three blocks of masses
`M_(a)`,
`M_(b)` and
`M_(c)` (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = (
`M_(a)` +
`M_(b)` +
`M_(c)` ) *a ................(equation 1)

Also it is given that
`M_(a)` does not move with respect to
`M_(c)`, which gives tension T is exerted on pulley by
`M_(a)` only, Hence tension T is

T =
`M_(a)` *a ..........(equation 2)

There is also also tension exerted by
`M_(b)`. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T =
`M_(b)`
`\sqrt{a^(2) +g^(2) }` ................(equation 3)

From equation 2 and 3, we get

`M_(a)` *a =
`M_(b)`
`\sqrt{a^(2) +g^(2) }`

Squaring both sides we get

`M_(a) ^(2)` *
`a^(2)` =
`M_(b) ^(2)` * (
`a^(2)`+
`g^(2)`)

`M_(a) ^(2)` *
`a^(2)` = (
`M_(b) ^(2)` *
`a^(2)`)+ (
`M_(b) ^(2)` *
`g^(2)`)

(
`M_(a) ^(2)` -
`M_(b) ^(2)`) *
`a^(2)` =
`M_(b) ^(2)` *
`g^(2)`

`a^(2)` =
`M_(b) ^(2)` *
`g^(2)`/(
`M_(a) ^(2)` -
`M_(b) ^(2)`)

Taking square root on both sides, we get acceleration a

a =
`M_(b)`*g/(
`\sqrt{M_(a) ^(2)-M_(b) ^(2)}`)

Hence substituting the value of a in equation 1, we get

F = (
`M_(a)` +
`M_(b)` +
`M_(c)` ) *(
`M_(b)`*g/(
`\sqrt{M_(a) ^(2)-M_(b) ^(2)}`))