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A 50.0-kg block is being pulled up a 13.0° slope by a force of 250 N which is parallel to the slope, but the block does not slide up the slope. What is the minimum value of the coefficient of static friction required for this to happen?
Answer is 0.292 but I dont know the working.
I split the vector of the 250N into x and y components and I also did it for the mass of the block(because its on a slope you can split Fg into Fgx and Fgy) and now i'm stuck.
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You don't need to split the 250N since it's already parallel to the slope.
The normal force (Fgy) produces a frictional force Ff = Fgy times mu.
Now look at Fgx. Is it directed in the same direction of Ff then the two forces must be added otherwise subtracted. You obtain the total opposing force which must be greater than 250N.
This gives an equation that you must solve for the friction coefficient mu.
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>>363835
this post is going to be a little bit confusing to a naive anon who assumes that your x axis is horizontal and your y axis is vertical
you skipped the part where you say
>first, set up your axes so that the x axis is parallel to the slope and the y axis is normal to it
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Your 250N is offset by
- gravity: 50g*sin(13) 110.23 and
- friction: 50g*cos(13)*x = 139.77
I got .293 lol.
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