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Mechanics Help

Hi,

I'm not getting some mechanics questions. I'm sure that I'm right... but the answer in the back of the book is different. I'll type it out for you guys

First Question:

"A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a horizontal force X. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle find X when the body is on the point of slipping:

a) up the plane.
b) down the plane."

I done a, but got it wrong, so I left out b, since I'm assuming I'm missing something.

I got 11.05 N, but the answer says 11.9N

The Second question is:

"Given that the resistances total 400N find the magnitude of the constant force needed to accelerate a car of mass 800kg from rest to 20m/s in 15 seconds."

Okay so I work out a, which is a=(v-u)/t=20/15=4/3

So F=ma=800x4/3=1066.67N

F+Resistance=Total Force=1466.77N, but the answer is 1470N apparently?? I'm confused.

Would like this to be resolved as soon as possible cos I've got alot of mechanics questions to do.

Thanks :)
 

Answers and Replies

Doc Al
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For the first problem: Show your work not just your answer.

For the second: They just rounded off to a reasonable number of significant figures; your answer matches.
 
Okay gonna be a bit difficult as you need a force diagram. But just imagine a box on an inclined plane 20 degrees to the horizontal, with mg acting down, Normal Reaction and Frictional Force.

Normal Reaction=19.6Cos20=18.4N

Therefore Frictional Force=18.4x0.2=3.68N

So Xcos20=3.68+19.6sin20

X=11.05N

And the answer in the back of the book is 11.9N??
 
Doc Al
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Careful. Since the applied force is horizontal, it affects the normal force.
Okay... how do I take that into account?? I don't know X in the first place? Give me a hint please :)
 
Doc Al
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Find the normal force (in terms of X) by analyzing the force components perpendicular to the incline surface.
 
Find the normal force (in terms of X) by analyzing the force components perpendicular to the incline surface.
Thanks for your help.
 

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