Question

The coefficients of static friction at the contacting surfaces are µA = 0.25, µB = 0.34, and µC = 0.39. The 100 kg roller (solid) and 40 kg tube (hallow) both have a radius of 150 mm. Determine the maximum amount of friction force that can act at point C before slippage occur. Answers are: 455 N 444 N 433 N 412 N 400 N 389 N 378 N 366 N

          The coefficients of static friction at the contacting surfaces are µA = 0.25, µB = 0.34, and µC = 0.39. The 100 kg roller (solid) and 40 kg tube (hallow) both have a radius of 150 mm. Determine the maximum amount of friction force that can act at point C before slippage occur. Answers are:
455 N
		
444 N
		
433 N
		
412 N
		
400 N
		
389 N
		
378 N
		
366 N
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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The coefficients of static friction at the contacting surfaces are µA = 0.25, µB = 0.34, and µC = 0.39. The 100 kg roller (solid) and 40 kg tube (hallow) both have a radius of 150 mm. Determine the maximum amount of friction force that can act at point C before slippage occur. Answers are: 455 N 444 N 433 N 412 N 400 N 389 N 378 N 366 N
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Transcript

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00:01 Okay, so this is the force diagram for the roller.
00:04 Let's take a look at a moment about point o.
00:06 Consider counterclockwise moment is positive, clockwise moment is negative.
00:10 And this will give us fa 1015 millimeter minus fc 10 150 millimeter is equal 0.
00:17 So i have fa is to fc, which means that the friction at point a is equal to the friction at point c.
00:25 So now let's apply force equilibrium for the roller.
00:27 So have net force along x direction is equal to p minus fc, minus n a, cosine 30 degree minus f a sine 30 degree is 0.
00:36 Na here is the normal force f .8, and p here is the force that was applied.
00:43 And along the wide direction, we have nc plus f a cosine 3 degree minus n a, sine 30 degree minus w, is good 0.
00:50 W here is the gravity of the roller.
00:55 So now let's take look at the force diagram for the tube.
01:00 So we'll consider the x direction of such model here, is along the surface and the wide direction is perpendicular to the surface.
01:10 So now let's take look at the moment about point d, which is the center of the circle here.
01:16 So consider counterclockwise moment is positive, clockwise moment is negative, and this will give us f8 times 150 millimeter minus fb times 150 millimeter is equal 0.
01:28 So we have f .a is equal fb, which means that the friction at point a is equal to friction at point b.
01:34 So now let's apply force equilibrium for this case.
01:37 So we have net force along x direction is equal to n .a minus fb minus wt sine 30 degree is 0.
01:44 So we have wt sine 30 degree is equal to na minus fb.
01:48 Wt here is the gravity for the tube.
01:52 And now on the y direction, we have nb minus fa minus wt cosine 3 degree is equal 3 degree is equal to nb minus fa.
02:01 So we assume that the slip occurs at point a.
02:04 Therefore, we have wt sin 30 degree is equal to n .a minus f a f .a., which is equal to na minus muana, since we assume that slip occurs at 0 .a.
02:15 So the friction at point a now is the static friction.
02:21 So if you do some arrangement here, eventually we have na is equal to wt 330 degree over 1 minus mua...
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