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anonymous
 one year ago
A 4000 kg truck is parked on a 15˚ slope. How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.
anonymous
 one year ago
A 4000 kg truck is parked on a 15˚ slope. How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer is 10000 N

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441558783042:dwFirst job is to determine the normal force, \(F_N\). Can you do that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ospreytriple yes, it is 39200 N

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, that's not it. You must resolve \(F_g\) into its components perpendicular and parallel to the incline.dw:1441559811875:dwThe normal force is equal in magnitude and opposite in direction to \(F_{g,x'}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry. **\(F_{g,y'}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[F _{g*y}\] + 37864.3 N

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ospreytriple Here is what I did:dw:1441560405546:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry for the delay...major computer issues on my end. I like your work, but there appears to be an inconsistency in the problem. The truck is in equilibrium, meaning that all forces in the x' direction sum to zero and all forces in the y' direction sum to zero. This leads to \(F_N = F_{g, y'} = 37,900\) N and \(F_f = F_{g, x'}=10,160\) N. I believe this is where the answer you gave earlier comes from. However, if this is true than the coefficient of friction cannot be 0.9. Rather, it would be\[\mu = \frac{ F_f }{ F_N } = \frac{ 10,160 }{ 37,900 } = 0.27\]???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm so confused... I don't really get it.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3dw:1441619539725:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\(\mu = 0.9\) is an upper limit. otherwise friction would start pushing things around :p if it's not moving, the the friction force on the truck simply equals the weight of the truck resolved down the 15 deg slope

jgomezgoni
 one year ago
Best ResponseYou've already chosen the best response.0Yes, friction force <= mu Normal force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 dw:1441648009736:dw

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1@vvvbb As others have tried to explain, the reason is as follows. Before motion starts, friction force is a variable force (called static friction) that is mobilized as required by the sum of external forces. The value can vary from 0 (in any direction) to a MAXIMUM of \(\mu\)N. If the resultant external force exceeds \(\mu\)N, then motion begins, and the static friction no longer holds. However, any net external force below \(\mu\)N will cause the mobilization of the friction up to the amount necessary to remain in static equilibrium. In the given problem, the force tending to move the vehicle down the incline is \(mg \sin(\theta)\), which equals 10156N (using g=9.81 m/s^2). The MAXIMUM frictional force available is \(\mu~ mg( cos(\theta))\)=34113N. Thus the required force to maintain static equilibrium is just 10156N, << 34113N.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much @mathmate ! For some reason your explanation made me understand. Thank you also to @IrishBoy123 !

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1You're welcome! Glad that it is resolved. :)
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