A 0.01 N ant is standing on the back of a couch inclined at 75* to the horizontal. What is the frictional force that is keeping the ant on the couch?!!
I am getting 0.0025 N as the answer but teacher says the answer is 0.01 N.
Please explain briefly on how that is the answer.
I am getting 0.0025 N as the answer but teacher says the answer is 0.01 N.
Please explain briefly on how that is the answer.
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If I understand this right, the angle between the couch and floor is 75° (so the couch is nearly vertical). If that's so, then I suspect the frictional force is closer to 0.01 than to 0.0025. (It can't be 0.01 exactly, but it might round off to that.) Let's see ...
I drew this diagram: There's a right angle AOB where the couch meets the floor. A is above O, and OB (the floor) is horizontal. The ant stands on the couch at point C, so ∠COB is 75° and ∠AOC = 15°.
Vertical vector DC = 0.01 N is the weight of the ant. (D is above C.) Draw vector DE perpendicular to OC, so E lies on the couch. ∠DCE = ∠AOC = 15°. Vector DE = 0.01 sin 15° is the normal force exerted by the ant on the couch, and vector EC = 0.01 cos 15° is the parallel force tending to make the ant slide down the couch.
The frictional force opposes the parallel force, equal in magnitude and opposite in direction. The frictional force is 0.01 cos 15° = 0.009659 N, which rounds off to 0.01 N.
Your teacher was right! The reason is because the angle is so steep. What else is going to hold the ant up there?
I drew this diagram: There's a right angle AOB where the couch meets the floor. A is above O, and OB (the floor) is horizontal. The ant stands on the couch at point C, so ∠COB is 75° and ∠AOC = 15°.
Vertical vector DC = 0.01 N is the weight of the ant. (D is above C.) Draw vector DE perpendicular to OC, so E lies on the couch. ∠DCE = ∠AOC = 15°. Vector DE = 0.01 sin 15° is the normal force exerted by the ant on the couch, and vector EC = 0.01 cos 15° is the parallel force tending to make the ant slide down the couch.
The frictional force opposes the parallel force, equal in magnitude and opposite in direction. The frictional force is 0.01 cos 15° = 0.009659 N, which rounds off to 0.01 N.
Your teacher was right! The reason is because the angle is so steep. What else is going to hold the ant up there?