Why Are My Mechanics Problem Solutions Incorrect?

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The discussion revolves around two mechanics problems where the user is confused about the correct answers provided in the textbook. For the first problem involving a body on an inclined plane, the user calculated the force required to prevent slipping but received a different answer than the book, prompting a request for clarification on how to account for the horizontal force's effect on the normal force. In the second problem, the user calculated the force needed to accelerate a car but found a discrepancy in the total force due to rounding in the textbook answer. Respondents suggested showing work for clarity and emphasized the importance of considering the components of forces acting on the inclined plane. The user seeks guidance on how to properly analyze the forces involved to resolve these discrepancies.
DigitalSpark
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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 a lot of mechanics questions to do.

Thanks :)
 
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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 going to 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??
 
DigitalSpark said:
Normal Reaction=19.6Cos20=18.4N
Careful. Since the applied force is horizontal, it affects the normal force.
 
Doc Al said:
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 :)
 
Find the normal force (in terms of X) by analyzing the force components perpendicular to the incline surface.
 
Doc Al said:
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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