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Static Friction Required to Keep the System from Moving (Two Boxes)
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[QUOTE="Steve4Physics, post: 6836430, member: 681522"] I know use of symbols is generally preferred, but here’s a numerical example which might help to explain what I mean. Sorry it’s a bit long. Let’s suppose: ##g = 10m/s^2## and ##\mu_s = 0.4## (I’ve chosen convenient values to make the arithmetic simpler). A consists of a stack of 10 metal discs, each with a mass of 1kg (weight of a disc = 10N). B’s mass is 2kg (weight = 20N). When the system is in equilibrium (not accelerating), the string’s tension equals the weight of B, i.e. T = 20N. The limiting static frictional force is ##\mu_s m_Ag = 0.4 \times10 \times 10 = 40##N. That means a tension of more than 40N is required to make A slip. So A does not slip; the frictional force acting on A simply equals the tension, 20N. Now we reduce the weight of A by removing a disc. The limiting static frictional force is now ##\mu_s m_A g = 0.4 \times 9 \times10 = 36##N. That means a tension of more than 36N is required to make A slip. So A still does not slip; the frictional force acting on A is still 20N. We keep reducing A’s weight by removing discs. Eventually we reach the point where A’s mass is such that the limiting frictional force (##\mu_s m_Ag##) equals the tension (20N). That means ##0.4 m_A g = 20##, so ##m_A= 5##kg. When there are only 5 discs left, A is on the verge of slipping - but the tension is still 20N. Now we remove another disc. The limiting static frictional force is ##\mu m_Ag = 0.4 \times 4 \times 10 = 16##N. For an instant ([U]but only an instant[/U]) the tension (20N) is greater than the limiting frictional force (16N) and the masses accelerate. This is a ‘discontinuity’, where the system changes from being in equilibrium to accelerating. Consider what happens next. A’s mass is now 4kg. The system accelerates. The acceleration depends on (amongst other things) the kinetic frictional force. But for simplicity and illustration purposes, let’s assume kinetic friction is negligible. If we imagine the string ‘straightened out’, we see the system is equivalent to a total mass of (4kg + 2kg) accelerated by B’s weight (20N). The acceleration is ##a(= \frac Fm)= \frac {20}6 ≈ 3.33 m/s^2## Considering A, the new tension (T’) is the accelerating force acting on A, so T’ (= ma) = ##4 \times 3.33## ≈ 13.3N. [U]The tension when accelerating (13.3N) is less than the tension when stationary (20N).[/U] A more accurate calculation, taking kinetic friction into count, would give the tension when accelerating closer to (but still less than) 20N. [/QUOTE]
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Static Friction Required to Keep the System from Moving (Two Boxes)
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