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I just recently started reviewing my old textbooks for fun (don't judge), and I'm having trouble wrapping my head around tension forces in this review question for the Laws of Motion chapter. diagram of problem

Before I start adding friction forces I just wanted to make sure I was on the right track with my answers so far, and there is no answer key for this question but it's a really interesting one. So,

If the inclined plane is frictionless and, the system is in equilibrium, find (in terms of m, g, Θ)

a) the mass M

b) the tensions T1 and T2

c) if the mass M is doubled (from answer a), the acceleration of blocks

d) if the mass M is doubled (from answer a), the tensions T1 and T2

My answers:

a) M=3mg(sinΘ)

b) T1=2mg(sinΘ) and T2=3mg(sinΘ)

c) a=(g/3)

d) T1=(8/3)mg(sinΘ) and T2=4mg(sinΘ)

Maybe it's a little much to ask but I'm hoping someone has the time to double check and let me know. And I promise this isn't a homework assignment, I'm just doing this for curiosity's sake and to stave off dementia lol

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  • $\begingroup$ The underlying conceptual question - how are forces resolved when tension is involved - is (in my opinion at least) sufficiently general to fit the site, but you may get a more positive reception if you abstract out the numerical details and ask how the forces should get resolved. $\endgroup$ Apr 10 at 19:12

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As to your answers:

a) wrong. $3mg\sin\theta$ is the weight force, not the mass, but that is, of course, just a minor slip. So $M=3m\sin\theta$

b) correct. correct.

c) wrong. The total (inertial) mass subject to acceleration is (now given $M=6m\sin\theta$) $$M_{tot}=3m+M =3m(1+2\sin\theta)$$ and the total accelerating force (i.e. after balancing out the equilibrium forces as in a)/b) ) is $$F_{tot}=(M/2)g=3mg\sin\theta$$ so $$a=F_{tot}/M_{tot}=g\frac{\sin\theta}{1+2\sin\theta}$$

d) wrong. $$T_1=2mg\sin\theta+2ma=2mg\sin\theta\left(1+\frac{1}{1+2\sin\theta}\right)=4mg\sin\theta\left(\frac{1+\sin\theta}{1+2\sin\theta}\right)$$ $$T_2=3mg\sin\theta+3ma=3mg\sin\theta\left(1+\frac{1}{1+2\sin\theta}\right)=6mg\sin\theta\left(\frac{1+\sin\theta}{1+2\sin\theta}\right)$$ the latter of which could also be obtained by $$T_2=Mg-Ma=M(g-a)=Mg\left(1-\frac{\sin\theta}{1+2\sin\theta}\right)=6mg\sin\theta\left(\frac{1+\sin\theta}{1+2\sin\theta}\right)$$.

Don't yield an inch to dementia! ;-)

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  • $\begingroup$ Thank you, I see my mistake. For my total Mass I used the 6msinΘ + 3msinΘ, and not 3m... Also yes, my first answer I simply wrote it down wrong, it was just 3msinΘ $\endgroup$
    – Ben Ji
    Apr 10 at 20:42

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