tbr ch 2 (forces)- static friction mistake?

[keikoblue2]

For question 39 from chapter 2 (forces), I think they got the answer choices mixed up? The question asks "which of the following graphs best represents the force due to static friction as incline angle increases for a redwood block atop a smooth rubber surface."

Graph C shows static friction increasing as incline angle (x) increases until a threshold, then it curves down.

Graph D shows the static friction increasing as incline angle increases, without a threshold. The graph continues in an upward slope.

Their answer was: "the static friction increases with the angle of inclination until the object breaks free and begins to slide down the surface. This can be seen in the graph in choice D. If choice C were true, then there would be no threshold value, because the maximum static friction force does not occur right as the object breaks free. This eliminates choice C."

I'm pretty sure they got the graphs of choices C and D mixed up, but I just wanted to double check with all the smart people here, since I'm paranoid. Thanks!

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[FeinMS]

Sounnds right to me. Maximum static friction occurs just before the box starts to move. This is shown in D not C.

 

[keikoblue2]

Sounnds right to me. Maximum static friction occurs just before the box starts to move. This is shown in D not C.

OH. OH I SEE. The static friction continues to the maximum, then it stops and kinetic friction occurs (which has a lower value). Thanks a lot!

 

[drechie]

if static friction is related to normal force, and normal force is proportional cos ( theta ) [Normal force is = mg * cos (theta)].... and as you increase the inclined plane theta increases and thus the value of cos (theta ) decreases.... wouldn't static friction decrease (in relation to increasing theta?)

 

[sazerac]

The maximum static friction is related to the normal force. If you push with a value less than this maximum, the object will push back with the same force (not the maximum possible force).

So as theta increases, the maximum static friction force is shrinking (just as you surmised), while the actual force exerted by static friction is increasing, and when those two values are the same, the block begins to slide.

 

[drechie]

The maximum static friction is related to the normal force. If you push with a value less than this maximum, the object will push back with the same force (not the maximum possible force).

So as theta increases, the maximum static friction force is shrinking (just as you surmised), while the actual force exerted by static friction is increasing, and when those two values are the same, the block begins to slide.

I don't see the mathematical logic between the equation --> Fstatic = mu(s) * mgcos (theta) <---- and your explanation. I know you are right but i must be foggy or something

 

[sazerac]

Fstatic is LESS THAN OR EQUAL TO mu(s) times mgcos(theta).

Alternately, the MAXIMUM Fstatic POSSIBLE is mu(s) times mgcos(theta).

Think about this for a second. You see some big rock. It takes 10,000 newtons of force to get it sliding. The maximum Fstatic is 10,000 newtons. You walk up to the rock, and lean on it with a force of 5 newtons. What force is it going to push back with? 10,000 newtons? Is it really going to start sliding towards you and crush you???

No, the Fstatic force is going to push back with exactly 5 newtons, and the rock won't budge. Fstatic <= FstaticMax = mu(s) times mgcos(theta).

 

[BerkReviewTeach]

Fstatic is LESS THAN OR EQUAL TO mu(s) times mgcos(theta).

Alternately, the MAXIMUM Fstatic POSSIBLE is mu(s) times mgcos(theta).

Think about this for a second. You see some big rock. It takes 10,000 newtons of force to get it sliding. The maximum Fstatic is 10,000 newtons. You walk up to the rock, and lean on it with a force of 5 newtons. What force is it going to push back with? 10,000 newtons? Is it really going to start sliding towards you and crush you???

No, the Fstatic force is going to push back with exactly 5 newtons, and the rock won't budge. Fstatic <= FstaticMax = mu(s) times mgcos(theta).

EXCELLENT!!! This is a point many people overlook. The value of static friction, like the value of tension and normal force, depends on the applied force. It is a reactionary force with equal magnitude but opposite direction of the applied force. The equation most people recall however, is Fstatic (max) = mus x N.

 

[ImDiene0412]

EXCELLENT!!! This is a point many people overlook. The value of static friction, like the value of tension and normal force, depends on the applied force. It is a reactionary force with equal magnitude but opposite direction of the applied force. The equation most people recall however, is Fstatic (max) = mus x N.

In the explanation, BR says the applied force is mgSin(theta), but since we are talking about an inclined plane, shouldn't the applied force be mgCos(theta)? Why is this [either scenario] true?

 

[TheLongGame]

Okay, gonna start fresh!

G = gravitational accel.
ã = some angle
M = mass
Us= coeff of static
Uk = coeff of kinetic
Fs = force of static
Fk = force kinetic

On an inclined plane at an angle ã, the X-Y components of "G" break down as follows:

Gx = Gsin(ã)
Gy = Gcos(ã)

The applied force that static friction must resist is proportional to the X-component of G. Of course we know that the proportional factor is the objects mass.

The confusing part is that you must use the Y-component of G to determine the force of the Normal and thereby determine the force of static friction at the given angle/mass/Us. The force "applied" to the block is due to MGxsin(ã), and is what the static force must overcome in order for the object to remain stationary.

When calculating the force of static friction, you are really calculating the maximum force the static friction can resist. So really: Fs -->Fs(max)

Any applied force less than this threshold is countered perfectly by a proportional value of Fmax --> Fs(applied) until the threshold value of Fmax is reached. Then we all know Fk and all of that happiness takes over.

Sorry for the non standard symbols....using a cell phone.


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[ImDiene0412]

THANK YOU! Very helpful!