# Min. downward force to keep object from slipping

• kevnm67
In summary: Good jobIn summary, the minimum downward force required to keep the box from slipping is 53 Newtons, calculated by setting the force of tension equal to the force of friction and using the equation Fn=Fg+Fa, where Fn is the normal force, Fg is the force of gravity, and Fa is the additional downward force. This is found by setting the coefficient of friction, 0.36, multiplied by the normal force equal to the applied force of 125 Newtons.
kevnm67

## Homework Statement

What is the min. downward force on the box that will keep it from slipping?
Coefficients are:
static= .36
kinetic= .26

the box is 30kg and there is a rope pointing -----> with 125N

F=ma
Ms n
Mk n

## The Attempt at a Solution

a= (Ms mg + Mk mg)/m = 6.1 m/s^2

Forces:
T= 125N
mg= -294N
n= 294N
Fs= .36
Fk= .26
a= 6.1m/s^2

F=ma -> F=125-.36-.26 = 124.4 ma= 30kg(6.1)= 183N

I am not sure if any of this is correct and am stuck moving on from here. If this is correct, would I divide F/ma? Thanks for any help or clarification you can offer

If the box isn't slipping that means it's not moving. What is the acceleration of a stationary box?

swells said:
If the box isn't slipping that means it's not moving. What is the acceleration of a stationary box?

The answer need to be in units of force and I tried two wrong answers; a= F/m = 4.2 m/s^2 and 0N. Here is the copied question:

"What is the minimum downward force on the box in the figure that will keep it from slipping? The coefficients of static and kinetic friction between the box and the floor are 0.36 and 0.26, respectively."

Well then you need to solve for a downward force that will make the force of friction equal to the force applied. You need to use the coefficient of static friction, as the box isn't moving.

Can you tell me why the stationary box has a different coefficient of friction than one that is moving?

As for solving, remember your net force equation. Since the box can't be accelerating Fnet=0.

Three steps: Free body diagram, net force equation, substitution.

Fnet= Ftension + Ffricton = 0
You can rewrite this as: Ftension=Ffriction

Don't forget that the force of friction is equal to the normal force applied by the surfaces it is in contact with, in this case the floor multiplied by the coefficient. Try writing the equation out that way and see if it helps. As I said, start with your free body diagram and label all the individual forces acting on it.

Static friction is harder to overcome due to bonds formed between the box and the surface.

Thanks, ill attempt it again

Alright, so your saying that the FBD has a -w, +n, +T. W= 294 and normal = 294+.36 = 294.36, and T= 125?

Sorry, it took me a bit to figure out what was going on with that last statement.

From what I interpret that's mostly right. I'll tell you how this is setup, because you seem not fully grasp what the diagram is for. I'll try describe it. Draw a box. Now given your setup you have four forces acting on it. You have the force of tension going in one direction, and the force of friction opposite to that (right and left). Vertically you have the normal force pushing up. Going down you both have the force of gravity, and the additional downward force.

Draw that out. This is what you use to setup your net force equation in each vector. Remember that you need to define a direction as positive. In this case up and right will be positive.

In your X direction you have: Fnet=Ff+Ft=0, which you can rewrite as Ff=Ft. Since in this setup Ff is negative. In the Y axis you have Fnet=0 as well, so Fn=Fg+Fa.

Now remember that Ff=(coefficient of friction)Fn. So let's put this together as Ft=(coefficient of function)Fn, and you have another equation that says Fn=Fg+Fa. That's plenty of hints. Put your algebra to use and solve for Fa.

Now you can see that Fn = Fg + Fa (force applied).

Lancelot59 said:
Sorry, it took me a bit to figure out what was going on with that last statement.

From what I interpret that's mostly right. I'll tell you how this is setup, because you seem not fully grasp what the diagram is for. I'll try describe it. Draw a box. Now given your setup you have four forces acting on it. You have the force of tension going in one direction, and the force of friction opposite to that (right and left). Vertically you have the normal force pushing up. Going down you both have the force of gravity, and the additional downward force.

Draw that out. This is what you use to setup your net force equation in each vector. Remember that you need to define a direction as positive. In this case up and right will be positive.

In your X direction you have: Fnet=Ff+Ft=0, which you can rewrite as Ff=Ft. Since in this setup Ff is negative. In the Y axis you have Fnet=0 as well, so Fn=Fg+Fa.

Now remember that Ff=(coefficient of friction)Fn. So let's put this together as Ft=(coefficient of function)Fn, and you have another equation that says Fn=Fg+Fa. That's plenty of hints. Put your algebra to use and solve for Fa.

Now you can see that Fn = Fg + Fa (force applied).

Thanks for the detailed response. I figured it out shortly before you responded:

.36(294+F)= 125
F=53

kevnm67 said:
Thanks for the detailed response. I figured it out shortly before you responded:

.36(294+F)= 125
F=53
That's what I got.

## 1. What is meant by "Min. downward force to keep object from slipping?"

The minimum downward force required to keep an object from slipping is the minimum amount of weight or pressure that needs to be exerted on the object to prevent it from moving or sliding. This is important in situations where objects need to be stabilized or held in place, such as on an inclined surface or during transportation.

## 2. How is the minimum downward force calculated?

The minimum downward force is calculated by taking into account the weight of the object and the coefficient of friction between the object and the surface it is resting on. The formula for calculating this force is F = μmg, where F is the minimum downward force, μ is the coefficient of friction, m is the mass of the object, and g is the acceleration due to gravity.

## 3. What factors can affect the minimum downward force required?

The minimum downward force required can be affected by several factors, including the weight and size of the object, the surface it is resting on, the angle of the surface, and the coefficient of friction between the object and the surface. Additionally, external factors such as air resistance or vibrations can also impact the minimum downward force needed to keep an object from slipping.

## 4. Why is it important to know the minimum downward force required?

Knowing the minimum downward force required is important for ensuring the stability and safety of objects in various situations. It can also help in determining the necessary equipment or materials needed to secure or transport objects, and can prevent accidents or damage caused by objects slipping or sliding.

## 5. How can the minimum downward force be increased?

The minimum downward force can be increased by increasing the weight or pressure on the object, or by increasing the coefficient of friction between the object and the surface it is resting on. This can be done through various methods such as adding weights on top of the object, using adhesive materials, or increasing the roughness of the surface.

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