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Correct me if I am wrong but here goes: Yea I get the x component to be 71N as well. The point at where the block is about to move is where the applied force is equal to the friction force. Friction = static coefficient * normal force I get 71 = static coefficient * 425 static coefficient = 71/425 = 0.167
The normal force, N\[N=425\] The force along the horizontal, Fx, is\[F_x=142\cos(\theta)\]
http://en.wikipedia.org/wiki/Friction#Dry_friction When it is just about to move, the force of friction equals the normal force times the coefficient of friction. So \[142 \cos(60)=71= \mu 425\]
So yes, you are correct.
no but the answer is .235
i know i am wrong somewhere
answer should be .235
ok how do u solve it
@Decart is correct- I did not take into account F_y, the fact that the force's y-component weakens the normal force. @ksaimouli , just subtract 142sin60 from the normal force.
i am confused
Okay- you push the block at an angle. This does 2 things- push the block ALONG (which is resisted by friction) and UP (opposite to gravity, so this reduces the block's force on the ground, as the force of the block DOWN is decreased, so it reduces the normal force the ground exerts on the block). It's as if you have a man pushing horizontally with force Fx and a cable with tension Fy pulling up on the block.
what is force of friction
The normal force changes, so the friction changes. @Decart , I'm not sure what you're doing, but Fy is defintely using sine This gives the right answer by the way.
instead of that can ii say 142cos60?
i got i got ohhhhhhhh it is simple guys
so friction is 71 by finding the Fn=302
thx u for helpinggggggggggggg me
i have only one question how did u get the equation Fn=mg-fy
i got u thx