Each of the plates A, B, and C have a mass of 10 kg. Plate A is on the bottom, p

Question

Each of the plates A, B, and C have a mass of 10 kg. Plate A is on the bottom, plate B is in the middle, and plate C is on the top. Plate A is being pulled to the right with a force of 15 N, Plate B is being pulled to the let with a force of 100 N, and Plate C is being pulled to the right with a force of 18 N. The coefficients of static and kinetic friction are 0.3 and 0.2, respectively, between each of the plates and between plate A and the ground. Determine the acceleration of each plate. Show all work and calculations

Explanation / Answer

follow this First of all, let's figure out whether there is slipping between Cand D. The maximum possible frictional force pushing C to the left due toits friction with B is (10+10)kg * .3 N/kg = 6 N. Knowing this,let's consider just the top two blocks. Therefore, the net force applied to C without taking intoaccount friction with D is at least 94 N to theright. The maximum value of static friction between C and D is 10kg * .3 N/kg = 3 N. Therefore, we will clearly have sliding here,since the 100 N force is much higher. Since the two blocks aresliding, friction between D and C pushes D to the right with aforce of 10 kg * .2 = 2 N, and pushes C to the left with the sameamount of force. Now we can update our initial diagram with thesefrictional forces: Now let's determine whether C and B are sliding. The notincluding the frictional force between C and B, the net forceon C is 98 N to the right (100 N - 2 N from the frictional force ofC and D). The maximum possible static frictional force is (10 kg +10 kg) * .3 N/kg = 6 N. Clearly 98 > 6, so C will be slidingpast B as well. Therefore, we again use the kinetic frictioncoefficient. So the frictional force on C due to B is 20 kg * .2N/kg = 4 N to the left, and the opposite pushes B to the right. Nowwe again update our diagram: All right, now all that's left is to determine whether B slidesagainst A. The maximum possible value of static friction between Band A is (10 kg + 10 kg + 10 kg) * .3 N/kg = 30 kg * .3 N/kg = 9 N.However, the force pushing B to the left is 15 N - 4 N = 11 N (the4 N is due to the frictional force between B and C). This isgreater than the maximum static frictional force, so B must besliding against A as well. Thus, the frictional force on B due to A is 30 kg * .2 N/kg = 6 Nto the right. So finally, we have: Adding up all these vectors, we see that: D experiences a force of 18 - 2 = 16 N to the left C experiences a force of 100 - 4 - 2 = 94 N to the right B experiences a force of 15 - 4 - 6 = 5 N to the left F = ma = 10a, and so a = F/10 D's acceleration = 16/10 = 1.6 m/s^2 left C's acceleration = 94/10 = 9.4 m/s^2 right B's acceleration = 5/10 = .5 m/s^2 left

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