A stationary 12.5 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.50 and of kinetic friction is 0.30. Show all work including units. A horizontal force of 15 N is applied to the object.

a. Draw a free body diagram with the forces to scale.

b. Determine the force of friction.

c. Determine the acceleration of the object.

When the object is at rest, there is a zero net force due the cancellation of the object's weight w with the normal force n of the table pushing up on the object, so that by Newton's second law,

∑ F = n - w = 0   →   n = w = mg = 112.5 N ≈ 113 N

where m = 12.5 kg and g = 9.80 m/s².

The minimum force F needed to overcome maximum static friction f and get the object moving is

F > f = 0.50 n = 61.25 N ≈ 61.3 N

which means a push of F = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.

So:

(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.

(b) Friction has a magnitude of 15 N because it balances the pushing force.

(c) The object is in equilibrium and not moving, so the acceleration is zero.

32: from largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages.

33: answer is c

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unbalanced force

explanation: