Answer:
The magnitude of the force F is given by
F = ([tex]M_{a}[/tex] + [tex]M_{b}[/tex] + [tex]M_{c}[/tex] ) *([tex]M_{b}[/tex]*g/([tex]\sqrt{M_{a} ^{2}-M_{b} ^{2}}[/tex]))
Explanation:
Given there are three blocks of masses [tex]M_{a}[/tex], [tex]M_{b}[/tex] and [tex]M_{c}[/tex] (ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ([tex]M_{a}[/tex] + [tex]M_{b}[/tex] + [tex]M_{c}[/tex] ) *a ................(equation 1)
Also it is given that [tex]M_{a}[/tex] does not move with respect to [tex]M_{c}[/tex], which gives tension T is exerted on pulley by [tex]M_{a}[/tex] only, Hence tension T is
T = [tex]M_{a}[/tex] *a ..........(equation 2)
There is also also tension exerted by [tex]M_{b}[/tex]. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T = [tex]M_{b}[/tex] [tex]\sqrt{a^{2} +g^{2} }[/tex] ................(equation 3)
From equation 2 and 3, we get
[tex]M_{a}[/tex] *a = [tex]M_{b}[/tex] [tex]\sqrt{a^{2} +g^{2} }[/tex]
Squaring both sides we get
[tex]M_{a} ^{2}[/tex] *[tex]a^{2}[/tex] = [tex]M_{b} ^{2}[/tex] * ([tex]a^{2}[/tex]+[tex]g^{2}[/tex])
[tex]M_{a} ^{2}[/tex] *[tex]a^{2}[/tex] = ([tex]M_{b} ^{2}[/tex] * [tex]a^{2}[/tex])+ ([tex]M_{b} ^{2}[/tex] *[tex]g^{2}[/tex])
([tex]M_{a} ^{2}[/tex] - [tex]M_{b} ^{2}[/tex]) * [tex]a^{2}[/tex] = [tex]M_{b} ^{2}[/tex] *[tex]g^{2}[/tex]
[tex]a^{2}[/tex] = [tex]M_{b} ^{2}[/tex] *[tex]g^{2}[/tex]/([tex]M_{a} ^{2}[/tex] - [tex]M_{b} ^{2}[/tex])
Taking square root on both sides, we get acceleration a
a = [tex]M_{b}[/tex]*g/([tex]\sqrt{M_{a} ^{2}-M_{b} ^{2}}[/tex])
Hence substituting the value of a in equation 1, we get
F = ([tex]M_{a}[/tex] + [tex]M_{b}[/tex] + [tex]M_{c}[/tex] ) *([tex]M_{b}[/tex]*g/([tex]\sqrt{M_{a} ^{2}-M_{b} ^{2}}[/tex]))
