Magnitude of the force of static friction

AI Thread Summary
The magnitude of the force of static friction on a block at rest on an incline is determined by the gravitational component acting down the slope, which is Mg sin(theta). While the maximum static friction force is given by mu_s1 * Mg cos(theta), this value is not the actual force acting on the block unless it is sliding. Since the block is at rest, the static friction force must equal the gravitational force component, thus confirming that the correct answer is choice c) Mg sin(theta). The discussions clarify that the static friction force can vary up to its maximum limit based on the conditions of motion. Understanding the distinction between maximum static friction and the actual force is crucial in solving such problems.
JessicaHelena
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Homework Statement



A block of mass M is at rest on an incline that makes angle theta with the horizontal. Given that the static coefficient of friction is mu_s1 and the kinetic coefficient friction mu_k, what is the magnitude of the force of static friction exerted on the block by the incline?

a) mu_s1 M g cos theta
b) mu_s1 M g tan theta
c) Mg sin theta
d) Mg / than theta

Homework Equations



F_f = mu * F_N
F_net x = 0
F_net y = 0

The Attempt at a Solution



Using that F_f = mu * F_N, since the block will have F_n = Mg cos theta due to F_net y = 0,I thought that F_f = mu_s1 * Mg cos theta, which is a). However, it is apparently choice c), and while I can see that using F_net x = 0, I don't get why a) would be wrong.
 
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JessicaHelena said:

Homework Statement



A block of mass M is at rest on an incline that makes angle theta with the horizontal. Given that the static coefficient of friction is mu_s1 and the kinetic coefficient friction mu_k, what is the magnitude of the force of static friction exerted on the block by the incline?

a) mu_s1 M g cos theta
b) mu_s1 M g tan theta
c) Mg sin theta
d) Mg / than theta

Homework Equations



F_f = mu * F_N
F_net x = 0
F_net y = 0

The Attempt at a Solution



Using that F_f = mu * F_N, since the block will have F_n = Mg cos theta due to F_net y = 0,I thought that F_f = mu_s1 * Mg cos theta, which is a). However, it is apparently choice c), and while I can see that using F_net x = 0, I don't get why a) would be wrong.
It would be wrong because μsMg cosθ is the maximum force of static friction not what is needed to keep the block at rest. As long as the component of gravity mg sinθ is less than that maximum value, the block will not slide. When mg sinθ exceeds that value, the block will slide. Conversely, if the block does not slide, the force of static friction must be mg sinθ.
 
JessicaHelena said:
Using that F_f = mu * F_N,
Careful! For static friction, mu*F_N is the maximum value of the friction force. The actual force could be anything up to that value.

Hint: The object is at rest.
 
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