figured out around 39 of them and this is litterally the last one and I have no clue how to do it...well i got an idea but none of my ideas have worked and ive submitted the answer 4 times and now im on my last submission...so can someone please help me.



8. [CJ6 4.P.060.]
The weight of the block in the drawing is 82.0 N. The coefficient of static friction between the block and the vertical wall is 0.500.

(a) What minimum force F is required to prevent the block from sliding down the wall? (Hint: The static frictional force exerted on the block is directed upward, parallel to the wall.)
N
(b) What minimum force is required to start the block moving up the wall? (Hint: The static frictional force is now directed down the wall.)
N

 

Looks like a trick question to me. The answer to both should be 82 N. There is no normal force pushing the block into the wall, yes? No normal force --> no friction, therefore you only have to counteract the weight of the block. No diagram though, so maybe I'm missing some element of the problem statement...

 

I need a picture

 

http://i29.photobucket.com/albums/c252/dup...ew/untitled.jpg



s2ki said the file was too large so there it is on photobucket

 

Okay, I'm not a mechanical engineer or physicist, nor do I pretend to be, however, I'm going to take a crack at it.

My answer to part (a) is half of 82 N (would that be considered 41 N?), because you already have a static frictional force that's directed upward, so half of the work is already completed. Though I'm curious what this "static frictional force" is, sandpaper maybe?

To (b), I'm going to say 82 N because you although you have the static frictional force directed upward, you still have to overcome the force of gravity in order to move the block upward.

Let us know when youy find out the answer.

Warren

 

nope...to omve it it would have to be greater then 82 newtons...and at the same time you gotta realize that when moving it up that the frictional force is working against you.......i dono no one i know has figured it out yet

 

Maybe it's me, but "The static frictional force exerted on the block is directed upward, parallel to the wall" is misleading, at least to me. Makes it seem that the force is working with you, not against.

Anyway, then my guess for (a) would be 82.5 N.

I would almost think it would also be 82.5 N plus the force of gravity, but then 82 N would be the force of gravity for this block, right?

For (b), my guess is anything greater than 81.5 because although the block weighs 82.0 N, the force is now working with you at the rate of .5 N, thus the block "feels" (for lack of a better word) like it weighs 81.5 N.

Warren

 

last time I took a physics course was about 12 years ago in high school,
but let me try this.

Friction=0.5xFxsin(40) (friction depends on F)

a) Friction is acting upwards so

F x cos(40)=82-0.5 x F x sin(40)+

Solve the equation for F. I belive 40 is in degrees so be careful when you enter that in the calculator where default is radians.

b) Friction is acting downwards,
F x cos(40)=82+0.5 x F x sin(40)

 

Having the picture helps just a tad.

Gaus is correct.

 

hmmmm...just for kicks...

(a) From the diagram, summing up forces in y-dir:

0 = Ffric + F*COS(40) - W

where,
Ffric = static friction force = F*SIN(40)*MU where MU is the static coeff of friction = 0.5,
W = weight of the object = 82 N.

Solving the equation for F, you get: F = W/[MU*SIN(40) + COS(40)]

(b) Since now the Ffric is acting downward, summing up forces in y-dir:

0 = - Ffric + F*COS(40) - W
everything else being equal, and solving for F:

F = W/[COS(40) - MU*SIN(40)]

Substituting numbers, (b) take more force than (a)...make sense?