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

first page

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.

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??

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

 

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

 

You do the rest.

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.

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?