3
$\begingroup$

There are two cases when force is acting on the square plate of side a in two different directions as shown in figure . enter image description here enter image description here

In first case the force is acting perpendicular to the plane that is inside the plane (screen).

In second case the force is acting perpendicular to the plane but it is now parallel to the plane (screen).

In both cases the torque is $a$ x $F$.

But only in first case the plate rotate not in the second case . What is the reason behind this.

$\endgroup$
  • 2
    $\begingroup$ In your case 2, (assuming it's sort of like a pet door), the hinge forces on the axis will provide a counter torque, it'll be distributed all through out the axis, you can't just zero all of them by taking the torque at any 1 point there. (if not all through out, where hinges are present, just 1 force isn't enough for equilibrium of the flap) $\endgroup$ – Rick Dec 9 '17 at 9:00
  • $\begingroup$ @Rick in case 2 , if we are writing the torque equation about one of the hinge point in that case there is torque of force F but not of hinge force (which is in opposite direction of F) as in that case $r$ x $F'$ will be zero as both are in same direction . $\endgroup$ – Koolman Dec 10 '17 at 6:12
  • $\begingroup$ Yes, that's why I said just 1 [hinge] force isn't enough for equilibrium of the flap. $\endgroup$ – Rick Dec 10 '17 at 6:14
  • $\begingroup$ No, the hinge force will also have a vertical component, which will give torque, the other component will prevent it from sliding away by balancing F. $\endgroup$ – Rick Dec 10 '17 at 6:25
  • 1
    $\begingroup$ Like I said, just 1 force may not be enough to keep it in equilibrium, in the updated figure your single hinge force set the flap in transitional equilibrium,but not rotational equilibrium, as the torques are still unbalanced, if there were at least 2 forces, they would adjust out to give both equilibrium, the vertical and horizontal components of both hinge forces will change to give about both equilibrium. For example, here's an FBD of a door. $\endgroup$ – Rick Dec 10 '17 at 11:46
1
$\begingroup$

If I understand correctly, the plate is hinged at the yellow line. (Equivalent to this, it could be pinned at any to points along the top edge.) This means that the edge is constrained along this line. No point along this edge can move. Consequently the plate cannot turn in its own plane, it can only turn around the hinge in a perpendicular plane.

The force in digram I is either wholly or partly perpendicular to the plane of the plate. (It could be a skew force, with components parallel and perpendicular to the plane of the plate.) The perpendicular component of force provides a torque about the hinge, which exceeds the torque due to the weight of the plate, and turns the plate.

In diagram II there is no component of force perpendicular to the plate, so there is no torque about the hinge. The applied force $F$ exerts a clockwise torque about the right-hand end of the hinge (black dot), but the weight of the plate, and reaction forces at other points along the hinge, create anti-clockwise torques which balance the torque due to $F$.

$\endgroup$

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.