Worked example

[stephen]

Hi,

 

I have a worked example of a question to do with forces and i just cant work how he gets to the answer? could you help me?

 

i will upload the example

 

Thanks

[stephen]

first page

[swansont]

What parts do you understand and what parts don't you understand? The force components from the drawing are one part, and the equations derived from them are another.

[stephen]

Right i cant upload the rest of the question so ill write it out

 

 

Continued

 

Fcos30 - 50 x sin30 = 1/3 (50cos30 + f sin30)

 

0.866 f -25 = 14.43 + 0.1667 f

 

f = 56.3 N

 

The force required to just move the block in the direction shown is 56.3 N

 

 

 

 

Any help will be much apreciated.

 

Stephen

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i follow it up to the last bit but then the answer throws me ?

As i wrote it goes from

 

0.866f -25 = 14.43 + 0.1667f

 

and somehow the above = 56.3 N??

[Fuzzwood]

This revolves down to not being able to solve for f:

 

0,866f - 0,1667f = 14.43 + 25

 

You do the rest.

[swansont]

i follow it up to the last bit but then the answer throws me ?

As i wrote it goes from

 

0.866f -25 = 14.43 + 0.1667f

 

and somehow the above = 56.3 N??

 

It's algebra at this point. Fuzzwood has given you the first step. Doing physics relies heavily on being able to solve algebraic equations, such as 1 equation with 1 unknown, and 2 equations with 2 unknowns.

[stephen]

first page

 

Thankyou for your help i now know how to get f.

 

However i was wondering if i mirrored the diagram (so i reversed the incline slope etc.) would the resolvents of force be the same as before?