Grade 11 Forces Help: Acceleration, Friction & Unit Test Prep

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    Forces Grade 11
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Homework Statement


Answer included:
94cdfba6cf2545f73ae8928b1220fb3d.png



Homework Equations


Ff = μFn, where mu can be kinetic or static.
Fnet = ma
Fg = mg
I think that's it.


The Attempt at a Solution


I got the first part, I need help with acceleration. There's no evidence that I tried the second part I guess, as I have no idea what to do. How can you get μk using the Ff static and μs? How are Ff kinetic and Ff static connected? To answer the second part, you need Ff kinetic which I have no idea how to get. Where did my teacher get the assumption that μk = 0.37? I have a unit test tomorrow on forces.

NEXT QUESTION:

Homework Statement


91747b002ad3fd1fda881d60ceb250d5.png



Homework Equations


cosine law
sin law
Fnet = ma
Ff = μFn, where μ can be kinetic or static.
Fg = mg
I think that's it.

The Attempt at a Solution


So I drew the FBD, and found that Fgx = 50.99 and Fgy = 239.89. I placed the vectors head to tail and found the resultant, which was 249.719 using cosine law. I then used sine law to find the angle opposite to the 160 vector, and subtracted 15 degrees to get the resultant degree going NW (I got 6.99 degrees). Divide 249 by the mass to find acceleration. I know I haven't incorporated friction, and I'm trying it again right now. The reason I'm posting it still is because I'm REALLY short on time and I want this answered in case I fail again.




Note: My teacher said he was in a rush when answering the questions, so there's a chance that they're wrong.

Thanks.
 
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Zarquad said:

The Attempt at a Solution


I got the first part, I need help with acceleration. There's no evidence that I tried the second part I guess, as I have no idea what to do. How can you get μk using the Ff static and μs? How are Ff kinetic and Ff static connected? To answer the second part, you need Ff kinetic which I have no idea how to get. Where did my teacher get the assumption that μk = 0.37? I have a unit test tomorrow on forces.
Those value have no (useful) relation. You have to treat μk as variable to get a formula for the acceleration. It is possible to plug in specific values of μk afterwards, this has been done with 0.37. You could use 0.36, or 0.01 or whatever, too.

NEXT QUESTION:
Looks fine so far. You can include friction at that point now.
 
So I could just leave my answer as (89 - (156.96 x μk))/16? If I sub in the 0.37 I'll get the right answer. So I could sub anything in for μk, as long as I identify what I'm assuming?

I wasn't able to figure out how to include it. :/ Could you show me please?
 
Zarquad said:
So I could just leave my answer as (89 - (156.96 x μk))/16? If I sub in the 0.37 I'll get the right answer. So I could sub anything in for μk, as long as I identify what I'm assuming?
Right. I would simplify the fraction a bit (if you have decimal values anyway, there is no point in keeping the denominator).

I wasn't able to figure out how to include it. :/ Could you show me please?
In which direction does friction point, and what is its magnitude? This allows to calculate the net force.
 
Alright, thanks.


The magnitude is equal to μk * Fn, where Fn = Fg. So it equals 60.92. But how do I find the direction?
 
I got a friend to help me with the second. My values were wrong because of a calculation error, but my process was right. For friction, you simply subtract the magnitude of friction from the resultant.