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Kinetic friction and acceleration

Asked Dec 18, 2007, 01:56 PM — 3 Answers
Two crates, of mass 74 kg and 108 kg, are in contact and at rest on a horizontal surface. A 731 N force is exerted on the 74 kg crate. The kinetic friction is .15

How do I find the acceleration?

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jiten55's Avatar
jiten55 Posts: 106, Reputation: 59
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#2

Dec 18, 2007, 05:41 PM
Total Normal Reaction = (74 + 108) g = 182 g

Total Friction Force = .15 * 182 * g

Net Force = 731 - .15 * 182 * g

Acceleration = Force/Mass

= (731 - .15 * 182 * g)/182

= (731 - .15 *182* 9.8)/182

= 2.546
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itsme_vipsdude's Avatar
itsme_vipsdude Posts: 29, Reputation: 3
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#3

Dec 19, 2007, 11:03 PM
The bodies are not lying on each oter but they ar just in contact....so the above work out is wrong according to me
jiten55 (Dec 20, 2007 04:09 AM): THEN WHY DONT YOU GIVE THE CORRECT ANSWER?   Source:
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tjsail's Avatar
tjsail Posts: 12, Reputation: 15
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#4

Dec 28, 2007, 10:28 AM
I agree with jiten55. It turns out that the force of friction is the same whether one is sitting on the other or if they are both sitting on the surface. You might quibble and say the way the equation is written it implies one is sitting on the other (contrary to the wording of the problem), but it comes out the same.

Total Friction Force = .15*(74 kg + 108 kg) g = .15*74 kg*g + .15*108 kg*g

There's some law from first year of algebra that says that's true.
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