# Why is the friction force on the bed opposite of the gravity [closed] So here is a simple problem and here is a diagram that I find found online. Why is the frictional force pointing downwards. I mean I get the correct answer if I follow this diagram and math checks out but intuitively. I thought the frictional force should be acting against the person falling off the bed due to downward gravity so I thought, frictional force should be pointing upwards NOT downwards. Can someone explain why? The questions asks for the smallest angle at which the person will begin to slip off. correct answer is provided in the bottom for your reference  ## closed as off-topic by user191954, GiorgioP, John Rennie, ZeroTheHero, Kyle KanosFeb 18 at 12:19

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• I doubt that you would get the correct answer if you took the frictional force as downward unless you made a sign error somewhere. But you'd have to post more details of your calculation. – noah Feb 16 at 22:40
• I think you're not taking into account the centripetal force acting on the man, needed to keep him rotating. en.wikipedia.org/wiki/Centripetal_force – Gert Feb 16 at 22:51
• @Gert so are you saying centripetal force results in pushing the man up ward and as a result you would have the frictional force acting in the downward direction? – CuriousJ Feb 16 at 23:00
• @AaronStevens: yes. Which way the friction force points depends on all forces, he can't just leave one type of force out. – Gert Feb 16 at 23:05
• @CuriousJ: the direction of the friction force can initially be chosen arbitrarily. Once you've got the force diagram, apply $\Sigma F_x=0$ and $\Sigma F_y=0$ to calculate all forces you didn't know yet. If you find that $F_f<0$ that means you chose the 'wrong' direction for $F_f$. For problems like this, the correct direction of $F_f$ must be calculated from all other applied forces. – Gert Feb 16 at 23:22

So, for low angular velocities, friction will point upward, and gradually to zero at a certain angular velocity (you can show that this angular speed is given by $$\omega^2 = g \tan \theta /R$$, where $$R$$ is the radius of the motion of the center of mass), and for higher angular velocities friction will point downward.