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

Answer:

The magnitude of the force F is given by

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

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/($\sqrt{M_{a} ^{2}-M_{b} ^{2}}$))