Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Position from joint of two rod where force should apply perpendicular

1 vote
39 views

Two rods of same length l but there masses and coefficient of friction with table are m, 2m and $\mu$, $2\mu$ respectively are joint at point O as shown in the figure. Find the position x from joint of two rod where force should apply perpendicular to the length of the rod to drag the rod with constant velocity .

For constant velocity the net force on each rod should be zero .
rod 1 : $F-\mu mg=0$
rod 2 : $F-2\mu (2mg) =0$
But how is this possible ?

asked May 5 in Physics Problems by koolman (4,116 points)
edited May 6 by koolman
I guess the rods are joined rigidly at O, not hinged. So you can treat them as a single rigid rod with non-uniform density and friction. The force is applied to one rod, not both. The rods move with constant velocity and they remain perpendicular to the force $F$ - ie they do not turn.

This means that the composite rod is in static equilibrium : in the frame of the COM it does not translate or rotate. So you can apply the TWO conditions for static equilibrium :
1. the resultant force on the object is zero
2, the resultant moment of force on the object is zero.
I could not understand how to solve as a single rigid rod with non-uniform density and friction.
You can deal with the friction as two separate forces, each acting at the centre of the respective uniform rod. Then you have 3 forces. The above conditions apply.
Is $\frac{-l}{3}$ make sense?

1 Answer

1 vote
 
Best answer

I presume that the rods are joined rigidly at O, not hinged. So they cannot rotate relative to each other, only as a single rigid body.

The force is applied to one rod, not both, and is distributed between them by internal forces which keep the structure rigid. In your attempt you have assumed that a force $F$ is applied to each rod, so that the total force on the structure is $2F$ instead of $F$.

Also you have assumed that the only forces on each rod are the applied force $F$ and the friction force on that rod. There are also forces and couples between the two rods which keep them aligned. If you analyse the forces on each rod using separate free body diagrams (FBDs) you must include forces which each rod exerts on the other. But it is simpler to model the whole structure as a single rigid body. Then you can ignore internal forces - ie those between the two rods.


There are 3 external forces acting horizontally on the whole structure :
1. applied force $F$ acting to the right at distance $x$ above point O
2. friction force $R_1=\mu mg$ acting to the left at distance $\frac12 L$ above O, and
3. friction force $R_2=4\mu mg = 4R_1$ acting to the left at distance $\frac12 L$ below O.

The rods move with constant velocity to the right and they remain perpendicular to the force $F$ - ie they do not turn. This means that the structure is in static equilibrium : in the frame of the COM it does not translate or rotate. So you can apply the two conditions for static equilibrium :
I. the resultant force on the structure is zero
II. the resultant moment of force on the structure about any point is zero.

Applying these conditions, taking moments about O, and substituting $R_2=4R_1$, we have
I. $F=R_1+R_2 = 5R_1$
II. $xF+\frac12 LR_2=\frac12 LR_1, xF=\frac12 LR_1-\frac42LR_1=-\frac32 LR_1$
Solving
$x=-\frac32 \frac{R_1}{F} L=-\frac{3}{10} L$

The force $F$ should be applied at a distance $\frac{3}{10} L$ below point O.


Note: Since this is a statics problem we might wonder why the question did not specify that the rods remain stationary, instead of moving with constant velocity.

When the rods move we can be sure that the friction forces on them are given by $R=\mu mg$. When the rods are stationary the friction forces can take any values up to $R=\mu mg$. The friction forces could be distributed in various ways and still fulfil the equilibrium conditions. See Is there a way to determine the exact value of static friction in this situation?

answered May 6 by sammy gerbil (26,096 points)
edited May 6 by sammy gerbil
...