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A chain with mass M hangs between two walls, with its ends at the same height. The chain makes an angle 0 with each wall, as shown in Fig. 2.31. Find the tension in the chain at the lowest point. Solve this in two different ways: (a) Consider the forces on half of the chain. (This is the quick way.) (b) Use the fact (see Problem 2.8) that the height of a hanging chain is given by y(x) = (1/a) cosh(œx), and consider the vertical forces on an infinitesimal piece at the bottom. This will give you the tension in terms of a. Then find an expression for a in terms of the given angle 0. (This is the long way.) M Fig. 2.31 Section 2.2: Balancing torques (a) Let F and T be the tensions at the wall and the lowest point, respectively. Looking at the y forces on half of the chain gives F cos 0 = looking at the x forces gives F sin 0 = T. These yield T = (M/2)g tan 0. (M/2)g, and (b) The slope of the chain is y' = sinh ax, which is approximately ax for small x. Consider a small piece that goes from –x to x. The weight is essentially P(2x)g. The upward component of the tensions at the two ends is essentially 2Ty' 2T(ax). Balancing these gives T = pg/a. Now let's find a. The length from the bottom, as a function of a general value of x, equals %3D | da = V1+ y/2 (1/a) sinh ax. cosh ax = (11) Therefore, M/2 = (p/a) sinh axo, where xo is the location of the wall. But the slope at the wall is sinh a0x (M/2) tan 0. Plugging this into the above T gives T = (M/2)g tan 0, in agree- ment with part (a). 1/ tan 0. So M/2 = p/(atan0) → p/a =

A chain with mass M hangs between two walls, with its ends at the same height. The chain makes an angle 0 with each wall, as shown in Fig. 2.31. Find the tension in the chain at the lowest point. Solve this in two different ways: (a) Consider the forces on half of the chain. (This is the quick way.) (b) Use the fact (see Problem 2.8) that the height of a hanging chain is given by y(x) = (1/a) cosh(œx), and consider the vertical forces on an infinitesimal piece at the bottom. This will give you the tension in terms of a. Then find an expression for a in terms of the given angle 0. (This is the long way.) M Fig. 2.31 Section 2.2: Balancing torques (a) Let F and T be the tensions at the wall and the lowest point, respectively. Looking at the y forces on half of the chain gives F cos 0 = looking at the x forces gives F sin 0 = T. These yield T = (M/2)g tan 0. (M/2)g, and (b) The slope of the chain is y' = sinh ax, which is approximately ax for small x. Consider a small piece that goes from –x to x. The weight is essentially P(2x)g. The upward component of the tensions at the two ends is essentially 2Ty' 2T(ax). Balancing these gives T = pg/a. Now let's find a. The length from the bottom, as a function of a general value of x, equals %3D | da = V1+ y/2 (1/a) sinh ax. cosh ax = (11) Therefore, M/2 = (p/a) sinh axo, where xo is the location of the wall. But the slope at the wall is sinh a0x (M/2) tan 0. Plugging this into the above T gives T = (M/2)g tan 0, in agree- ment with part (a). 1/ tan 0. So M/2 = p/(atan0) → p/a =

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Below is the solution. For the (b)question, Would you like to explain why the slope at the wall is sinh alpha0x = 1/tantheta? Thanks.

A chain with mass M hangs between two walls, with its ends at the same height. The chain makes an angle 0 with each wall, as shown in Fig. 2.31. Find the tension in the chain at the lowest point. Solve this in two different ways: (a) Consider the forces on half of the chain. (This is the quick way.) (b) Use the fact (see Problem 2.8) that the height of a hanging chain is given by y(x) = (1/a) cosh(œx), and consider the vertical forces on an infinitesimal piece at the bottom. This will give you the tension in terms of a. Then find an expression for a in terms of the given angle 0. (This is the long way.) M Fig. 2.31 Section 2.2: Balancing torques (a) Let F and T be the tensions at the wall and the lowest point, respectively. Looking at the y forces on half of the chain gives F cos 0 = looking at the x forces gives F sin 0 = T. These yield T = (M/2)g tan 0. (M/2)g, and (b) The slope of the chain is y' = sinh ax, which is approximately ax for small x. Consider a small piece that goes from –x to x. The weight is essentially P(2x)g. The upward component of the tensions at the two ends is essentially 2Ty' 2T(ax). Balancing these gives T = pg/a. Now let's find a. The length from the bottom, as a function of a general value of x, equals %3D | da = V1+ y/2 (1/a) sinh ax. cosh ax = (11) Therefore, M/2 = (p/a) sinh axo, where xo is the location of the wall. But the slope at the wall is sinh a0x (M/2) tan 0. Plugging this into the above T gives T = (M/2)g tan 0, in agree- ment with part (a). 1/ tan 0. So M/2 = p/(atan0) → p/a =
Transcribed Image Text:A chain with mass M hangs between two walls, with its ends at the same height. The chain makes an angle 0 with each wall, as shown in Fig. 2.31. Find the tension in the chain at the lowest point. Solve this in two different ways: (a) Consider the forces on half of the chain. (This is the quick way.) (b) Use the fact (see Problem 2.8) that the height of a hanging chain is given by y(x) = (1/a) cosh(œx), and consider the vertical forces on an infinitesimal piece at the bottom. This will give you the tension in terms of a. Then find an expression for a in terms of the given angle 0. (This is the long way.) M Fig. 2.31 Section 2.2: Balancing torques (a) Let F and T be the tensions at the wall and the lowest point, respectively. Looking at the y forces on half of the chain gives F cos 0 = looking at the x forces gives F sin 0 = T. These yield T = (M/2)g tan 0. (M/2)g, and (b) The slope of the chain is y' = sinh ax, which is approximately ax for small x. Consider a small piece that goes from –x to x. The weight is essentially P(2x)g. The upward component of the tensions at the two ends is essentially 2Ty' 2T(ax). Balancing these gives T = pg/a. Now let's find a. The length from the bottom, as a function of a general value of x, equals %3D | da = V1+ y/2 (1/a) sinh ax. cosh ax = (11) Therefore, M/2 = (p/a) sinh axo, where xo is the location of the wall. But the slope at the wall is sinh a0x (M/2) tan 0. Plugging this into the above T gives T = (M/2)g tan 0, in agree- ment with part (a). 1/ tan 0. So M/2 = p/(atan0) → p/a =
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