As an aid in working this problem, consult Concept Simulation 4.4 The drawing sh

Question

As an aid in working this problem, consult Concept Simulation 4.4 The drawing shows a large cube mass 42 kg being accelerated acros5 a horizontal frictionless surface by a horizontal force P. A 5mall cube ma55 5.0 kg is in contact with the front surface of the large cube and will slide downward unless P Is sufficientiy large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that P can have in order to keep the small cube from silding downward? 0.71 Frictionless

Explanation / Answer

The horizontal force P is pushing (horizontally, of course!) on the small cube, and the small cube is prevented from sliding down the face of the large cube by friction force, which is F(u) = P * 0.71, which acts upward.
And the force acting downward on the small block is
F(d) =m * g = 5 * 49.
For the small block not to slide down,
F(u) = F(d), or P * 0.71 = 5 * 9.8

Now you solve for P:

P * 0.71 = 5 * 9.8

P = 5 * 9.8 / 0.71

P = 69.1 N

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