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Friction problem

So in this problem you're asked to find the magnitude as well as direction (upwards is positive) of the friction force on the block.

The block has a mass $m$ and is not moving in any direction.

If you consider the case where $\theta = 0$ then the force of friction obviously has to point upwards to keep $mg$ from moving the block downwards.

But for $ 0 < \theta \leq \pi/2$ the direction of the friction force depends on the relationship between $mg$ and $F$ which we know nothing about.

The magnitude will be either   $mg - F\sin\theta $      or    $F\sin\theta - mg$

depending on the direction but how can i decide the latter?

(the $\mu_k$ in the figure is for another problem)

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4 Answers 4

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The equation of equilibrium in the vertical direction is $F\sin \theta +F_f-mg=0$, where $F_f$ is the friction force (if its sign is positive, it is directed upwards), so $F_f=mg-F\sin\theta$. However, if $|mg-F\sin\theta|>\mu_s F \cos\theta$, the problem does not have a solution (the block cannot be in equilibrium).

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The friction force depends on the normal force $N$ of the wall pushing against the block. Newton's third law tells us that normal force is equal and opposite in direction to the force the block exerts on the wall, which is dependent on $F$ and $\theta$. The direction of the friction force will be opposite the direction of motion.

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If you change the angle $ \theta $ of the applied force $ \vec F $, it will only affect the magnitude of the friction force $ \vec F_f $, as you say (which is proportional to the normal force $ \vec N $ of the surface = wall), but the direction will not change: it'll always be vertical. It will only change the sense of the frictional force:

  • If the block goes up the friction force will be vertical pointing down

  • If the force F is not large enough, the block will fall, so $ \vec F_f $ will be vertical pointing up

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Always remember the following statements:

  1. Direction of Static friction is always opposite to the direction of the net force acting on the body.
  2. Direction of Kinetic friction on that body is always opposite to the relative speed of that body with respect to the other body.

Now coming to your question. If $\theta$ in the clockwise direction between 0 to $\pi/2$ radians, the net force is in the upward direction, therefore static friction will be in the downward direction(assuming it does not exceed $\mu F\cos(\theta))$.

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