Two crates, of mass m1 = 70 kg and m2 = 115 kg, are in contact and at rest on a horizontal surface (Fig. 4-54). A 620 N force is exerted on the 70 kg crate. The coefficient of kinetic friction is 0.15.
Calculate the force that each crate exerts on the other.
2006-10-05 15:00:36 · 2 answers · asked by Diane W 1 in Science & Mathematics ➔ Physics
Seems to me that since they are touching, they act as one (assuming that the force is exerted in line with the two crates - otherwise, the force will vary with the angle). If the thinking is that the smaller box "uses" some of the friction before the second box moves, this is incorrect because they act as one, connected unit. There is no time that the first can move before the second.
So, the answer is 620 N.
2006-10-05 15:18:38 · answer #1 · answered by lx470_98 1 · 0⤊ 1⤋
I don't agree with the previous responder.
Think of it as follows:
The 620 N force is resisted by 4 things:
1. the friction between m1 and the surface (which you can calculate, using its mass to calculate the force of gravity on it, and multiplying that by the coefficient of friction).
2. the friction between m2 and the surface
3. the inertia of m1 (i.e., its resistance to being accelerated, based on the classic formula, F = ma)
4. the inertia of m2
You know the amount of force used up by items 1 and 2. So you know how much force remains to accelerate the two masses. And you know the total mass, so you can solve for acceleration.
Finally, the force on m2 is equal to the force calculated in 2. above, plus the force needed to accelerate it at the rate of acceleration calculated in the preceding paragraph.
Come to think of it, it will turn out that the force on m2 is equal to 620 N times 115 / (70 + 115). I just realized that all of the math above leads to the conclusion that the force on m2 is based on its proportion of the total mass. But the above analysis may help you to see WHY that's the answer.
2006-10-05 16:29:36 · answer #2 · answered by actuator 5 · 3⤊ 0⤋