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I am confused between How forces work when applied in singular linear direction vs Force Applied to a continuous mass about an axis(example to rotate it).

Example 1: Consider an object of mass $m$ which is lying on a table with friction coefficient $k$, so the minimum force $F$ required to move the object will be $F = Mgk$

Example 2: Consider a similar example with a slight change.

Imagine the object of mass $M$ is for example a door, held straight with hinges. The Hinges also have a friction coefficient $k$, will the minimum force required to swing the door open be equal to the force we found in Example 1?

Also, I cannot understand how to calculate and what to consider to find the minimum force required to move an object along its axis or to move it in one direction and set it in ideal motion.

Imagine the object has same dimensions and properties in both examples.

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  • $\begingroup$ Go to the door in your room and try and open it by applying a force near the handle. Now apply the same force near the hinges... can you move it? For a rotation, the distance at which you apply the force from the axis is critical. You get a rotation from a force applied directly on an axis (try pushing on the hinges themselves...) $\endgroup$ – Jonathan R Swift Jul 5 at 17:36
  • $\begingroup$ @Jonathan thanks for your response , Yeh I understand applying force near the hinge will be more , the question is how much "minimum" / "starter" force should be applied no matter at "x" distance from the pivot point ? How much to set it in motion ? At "x" distance ? $\endgroup$ – german_wings Jul 6 at 2:30
  • $\begingroup$ Right, but the problem you have is your statement "the Hinges also have a friction coefficient $k$". -You haven't included a term for the radius of the hinge sliding surface itself - if you are using an idealised 'axis', then you need to specify the resistance as a torque. $\endgroup$ – Jonathan R Swift Jul 6 at 12:28
  • $\begingroup$ @JonathanRSwift Thanks , I am even more confused now , Idealized Axis ? The Hinge Do not provide a Axis for the door to rotate ? Is not the Farthest Point on the door has its centre of Rotation as the Hinge ? Then W of the door is its Radius. I am trying to conduct a practical experiment in my door. But maybe if you can Calculate the Minimum force required let me know in the Answer Section. Thanks $\endgroup$ – german_wings Jul 6 at 15:25
  • $\begingroup$ The hinge itself must have some size - say the hinge has a radius $r$, then the friction has a torque of $M*g*k*r$. $\endgroup$ – Jonathan R Swift Jul 6 at 18:01
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Take a look at the illustration below, which is how I understand your question. You're asking, what would be the minimum value for the blue arrows in order to make each object move? 3D view of problem

First, let's look at the simpler case - the red box.

I don't think this needs another diagram - you've rightly stated in your question that if there is a coefficient of friction between the red box and the floor of $k$, then the resistive force to motion will be equal to $k$ multiplied by the reaction force, which can be calculated from the mass of the box and the acceleration due to gravity. i.e. $F_{min}=Mgk$

Now, let's take a look at the green door. Let's call the width $x$.

The torque being applied is equal to $F*x$ but what's the reaction force?

First, we need to look how much force is being applied to the bearing surfaces of the hinge. There is a contact face at the bottom of the door, providing an upwards reaction equal to $M*g$, as before (to ensure vertical equilibrium), but because the centre of mass of the door is away from the hinge, there is a clockwise moment which must be resisted by the hinges, also. This clockwise moment is equal to $M*g*\frac{x}{2}$

The two red arrows show the moment reaction forces - the distribution between these is not critical - we must simply note that, combined, they sum to $M*g*\frac{x}{2}$.

So, we have some forces, and by multiplying by $k$, we can get some resistance to movement forces, but, there is one thing missing to tell us how much resistive torque this can generate - the radius at which it is acting.

Front view of door

Looking at the image below, the we will call the hinge pin radius $r_i$, and the hinge body radius $r_o$. Clearly, then the friction acting on the pin from the moment reaction is applied at the surface of the pin, at radius $r_i$. To find the effective radius at which the friction on the bottom face of the door is acting, we must find the radius where the surface area can be split into two equal halves. Let us call the unknown radius $r_u$. This gives us $\left(\pi{r_u}^2-\pi{r_i}^2\right)=\left(\pi{r_o}^2-\pi{r_u}^2\right)$ which can be solved to give $r_u=\frac{\sqrt{r_i + r_o}}{\sqrt{2}}$

The total resistive force due to friction in this example is therefore $\frac{Mgk\sqrt{r_i + r_o}}{\sqrt{2}}+\frac{Mgkxr_i}{2}$

Divide through by X, and you have your final answer for $F_{min}=Mgk\left(\frac{\sqrt{r_i + r_o}}{x\sqrt{2}}+\frac{r_i}{2}\right)$.

Hinge action radii

Clearly, the values of $r_i$ and $r_o$ will be comparatively small when compared to $x$, so you can assume that $\left(\frac{\sqrt{r_i + r_o}}{x\sqrt{2}}+\frac{r_i}{2}\right)<1$

That is to say, the force required to move the door, is smaller than the force required to slide the box.

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  • $\begingroup$ Thank You for your answer I , your solution "To Consider Friction of Hinges" works pretty well , yesterday , I put my door to test , I noticed the pulling force required to push my door is between 500gms to 600gms , while your equation yields a value 1.6Kgs , which is very less then had my door been lying on the floor without hinges , so it sums up "The Force required to Push a Object on the ground is far more then the object rotating along the pivot provided the friction between the object and its axis must be less." , Thank You $\endgroup$ – german_wings Jul 9 at 7:34
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The minimum force required to swing the door open would not be equal to the minimum force to push the object off the table.

Pushing the door that is attached to the door hinge would create different minimums of force required to push the door open as you could push the door at many different areas. In order to push a door open, you would need a certain amount of torque instead of force; increasing the lever arm decreases the amount of force needed to push against the friction of the hinges of the door.

A really cool place to start understanding what minimum forces you need to push an object along its axis is looking into dynamics- specifically studying movement of gears could help! Another way to understand object movement along its axis is through understanding the object's movement as a property of its moment of inertia.

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  • $\begingroup$ thanks can you present a proof for your answer ? $\endgroup$ – german_wings Jul 7 at 2:24
  • $\begingroup$ Yes! I will add it as a different answer. $\endgroup$ – Elizabeth Karlovics Jul 7 at 12:15
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Solution

Example 1: The object on the table would require a minimum force of $Mgk$.

Example 2: The simplest way to the able the amount of force required to move the door hinge that has friction would be to look at a cross section of the component. The cross section would be a circle with two forces being applied. These force would cause torque which will be essential to find out the minimum amount of force required to move the door. The assumption will be that friction will be caused by the axle, and by the knuckle onto the knuckle on top of it. The door and knuckle act as one body, and the pin and other knuckle acts as a separate body. Below is a FBD of the simplified situation: enter image description here

Step 1

To see what minimum applied force is necessary on the door to surpass the friction, you must look at the torques.

$$\tau_{applied} = \tau_{friction}\\ \tau_{applied} = \tau_{f_1} + \tau_{f_2}$$

Here, $\tau_{f_1}$ is the frictional torque due to the pin on knuckle force. The normal force we will look at will be at the point of the frictional force. $\tau_{f_2}$ is the frictional torque due to the knuckle on the other knuckle.

Step 2

Next, we plug in the equation $\tau = Fr\sin(\theta)$ find the torques. Before this though, we must find the radius that $F_{f_2}$ is being applied at. The term $w$ is the width of the door, $r_P$ is the radius of the pin, and $r_k$ is the radius of the knuckle.

$$r_2 = r_k - ((r_k-r_p)/2)$$

Now knowing the radius for $r_2$, where one of the average frictional forces is, we can plug into the equations.

$$ F_{applied}w\sin(\theta) = F_{f_1}r_p\sin(\theta_1) + F_{f_2}r_2\sin(\theta_2)\\ F_{applied}w = F_{f_1}r_p + F_{f_2}r_2 $$

Here, $F_{f_1}$ is the frictional force due friction between the pin and knuckle, and $F_{f_2}$ is the frictional force due to the friction between the knuckle and other knuckle. $\theta_1$ and $\theta _2$ are 90 degrees, and $\sin(90)=1$.

Step 3

Plug in frictional force equations. $$F_{applied}w\ = kF_Nr_p + kg(m_d+m_k)r_2$$

In this, $m_d$ is the mass of the door, and $m_k$ is the mass of the knuckle. The mass of the pin is unnecessary here. We will assume the frictional coefficient is the same for all the components.

Step 4

Simplify and solve.

$$F_{applied} = \dfrac{kF_Nr_p + kg(m_d+m_k)r_2}{w}$$

Assuming similar dimensions and properties given in the first example: $$F_{applied} = \dfrac{kF_Nr_p + Mgkr_2}{w}$$

Conclusion

To swing the door open, the torque would be used to find the necessary force needed to apply. This means that even after assuming similar condition, pushing the door would require taking the radius where the force is applied into consideration, along with the radius where the average frictional force is. The pin would also cause friction against the knuckle, which would be needed to be taken into consideration.The minimum force required to swing the door open will not be equal to the force we found in Example 1.

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  • $\begingroup$ Hey , I was looking at your solution , It does make sense to me , you didn't mention what is Fn in your equation at Step 3 , I read your solution at least 10 times now , but could not pick up what is Fn , could you point me ? Please , Thank You in advance. $\endgroup$ – german_wings Jul 9 at 7:29
  • $\begingroup$ Thanks! Fn would be the normal force of the pin on the knuckle that would cause the friction to occur $\endgroup$ – Elizabeth Karlovics Jul 9 at 16:23

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