0
$\begingroup$

I am working on the following problem:

Hibbeler, Dynamics: Ch13, Problem 64

Note that 0.5*N is the force due to friction pictured in the FBD.

When I try to solve this, my intuition is to draw the friction force upwards rather than downwards. If the "cushion" wasn't spinning, then the guy would slide downwards, and the friction force would be upwards. Depending on how quickly he is spinning, the friction force COULD be upwards, but how do we know with this type of problem?

I tried solving the problem my way, and I do get a different answer.

$\endgroup$
1
  • $\begingroup$ Add friction in whatever direction you think is right, and if the result is negative number, then reverse the sense. $\endgroup$ Feb 21, 2016 at 20:02

4 Answers 4

1
$\begingroup$

You can decide on the direction of the frictional force in the following way.

When $\theta = 90^\circ$ the frictional force must be upwards to balance the weight of the man.

As you decrease the angle $\theta$ the upward frictional force must decrease because eventually there will be an angle of inclination where no frictional force is needed to keep the man rotating. This is sometimes called the "ideal" banking angle given by $\tan \theta_i = \dfrac {v^2}{rg}$.

Decreasing the angle further must mean that the frictional force now point down the slope.

Eventually that frictional force is just sufficient to keep the man from sliding up the slope. That is then the minimum angle of the slope required by the questioner.

$\endgroup$
1
  • $\begingroup$ While many submitted helpful answers, this one makes the most intuitive sense to me. Thank you. $\endgroup$ Feb 20, 2016 at 9:23
0
$\begingroup$

How I would solve the problem is - Take axes along and perpendicular to the bed. From the frame of reference of rotating axis, draw the forces on the man. The gravitational, centrifugal and frictional force. Now for frictional, as you correctly said, the will be two cases :

  1. The component of centrifugal force in the upward direction is higher that the gravitational force's downwards. In that case the frictional force is downwards ( All ups and downs are along the bed ).

  2. When the component of centrifugal force in the upward direction is lesser that the gravitational force's downwards. Then the frictional force will be upwards.

    I hope that I am clear enough. :)

$\endgroup$
0
$\begingroup$

Force of friction is always in opposite direction to - relative direction of relative motion, or relative direction of attempted motion, that is why it is called friction.

$\endgroup$
7
  • $\begingroup$ My question is how to determine the direction of motion when the parameters that determine said motion are left as variables. $\endgroup$ Feb 20, 2016 at 5:52
  • $\begingroup$ The question title is "How to determine the direction of friction force on FBD?" $\endgroup$
    – kpv
    Feb 20, 2016 at 5:55
  • $\begingroup$ Yes, but answering that question requires knowing something about the direction of motion for this problem. Did you read the problem statement? $\endgroup$ Feb 20, 2016 at 6:53
  • $\begingroup$ @kpv I think the wording should be that friction opposes relative motion because can increase the speed of a body. $\endgroup$
    – Farcher
    Feb 20, 2016 at 7:09
  • $\begingroup$ I am not aware of any situation where friction increases speed. $\endgroup$
    – kpv
    Feb 20, 2016 at 7:11
0
$\begingroup$

The thing is you don't know, in general, we can narrow the direction down to two directions in opposite directions , in the equations, put f as a signed variable and you're good to go, if it comes out negative, then the assumed direction was wrong. In this particular question, the question demands logical thinking that in which case the angle would be smaller, so, start out with theta=0 {in your mind} and start slowly increasing it and mentally trying to figure out the direction of friction, ex - when theta=0.001 rad, it would be planar almost and almost circular motion, so, friction would act down {to provide centripetal force} , and as the angle increases, friction would start to act up

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.