box on inclined plane with friction Q, thanks (1 Viewer)

[shmitler]

 

[Drdusk]

View attachment 27327

This is your standard resolve the forces type of question. Let x be the direction parallel to the slope and y be perpendicular to the slope. Resolve this way we get

 

[shmitler]

sorry forgot to mention this, but the answer on the exam is 351.4 N
little bit confused by that

 

[Drongoski]

Is the answer about 349.5 Newton (using g = 9.8) or 351.53 N (g = 10)??

 

[shmitler]

I assumed they used g=10, yeh the answer is 351.4

 

[shmitler]

could you explain what you did to get that answer, would be really helpful
Thanks

 

[Drongoski]

Unfortunately I don't know how to provide diagrams ( which would have helped greatly). So I need to explain as best I can.

The weight 40g, can be resolved into 2 components: 40g sin(30) down slope and 40g cos(30) perp to (and against) the slope.
The force P can be resolved into 2 comps: P cos(30) up slope and P sin(30) perp and against slope.
.: total normal reaction (perp to slope) = 40g cos(30) + P sin(30)
.: the friction = 0.2 x the normal reaction
The component of P up slope must be just enough (equal to) to overcome the component of the weight down the slope (40g sin(30) )
as well as the friction

You end up with:
(sqrt(3)/2 - 0.1) x P = 20g + 4sqrt(3)g

Hope you can follow my explanation.

 

[shmitler]

thanks a lot, appreciate it

 

[Drdusk]

Take note kids, don't rush questions like me lol. Anyway by using g = 10 and flipping some signs in mine you'll get the answer, so still learn the method.

 

[shmitler]

thanks for your time anyway man