# Force of static friction in an FBD

• noleguy33
In summary, the conversation discusses a problem involving a donkey pulling a cart with different coefficients of friction. The first part involves the donkey pulling the cart with zero acceleration, while the second part involves the donkey pulling the cart with an acceleration of 0.24[m/s2] without slippage. The objective is to find the force exerted by the donkey on the cart and the force of static friction on the donkey by the surface. The conversation also discusses the use of Newton's 2nd law and drawing a free body diagram to find the solutions. Finally, it clarifies the concept of friction, which can act in either the direction of movement or against it depending on the situation.
noleguy33

## Homework Statement

The first part is a donkey is pulling a cart with zero acceleration and I've found the coefficient of kinetic friction in the bearings and the wheel, and the coefficient of static friction of the donkey in the ground. Part two I'm having problems with-

This time, a the donkey is pulling hard (almost starting to slip) on the same cart. The cart and donkey are accelerating at 0.24[m/s2] to the left without slippage.

Find-
a) The force by the donkey on the cart
b)The force of static friction by the surface on the donkey.

Newton's 2nd law

## The Attempt at a Solution

My first step is always to draw a FBD.

This is how I found Fd/c

Ffk - Fd/c = m(-a) /// set up that way because it's moving towards the negative x-axis

I used the same coefficient for from the first part of that problem, I got an answer of 2065[N].

I'm having an issue with the second part.

I drew a FBD for the donkey and came up with for the x-axis-

Ff(static) - Fc/d = m(-a) and I believe that gets me the right answer(or at least a common answer among classmates).

I just don't know why. Given this, the Ff(static) would be larger than Fc/d, so how exactly is the donkey moving? Thanks!

My only thought is that the Ffs is a force of the Earth on the donkey?

noleguy33 said:

## Homework Statement

The first part is a donkey is pulling a cart with zero acceleration and I've found the coefficient of kinetic friction in the bearings and the wheel, and the coefficient of static friction of the donkey in the ground. Part two I'm having problems with-

This time, a the donkey is pulling hard (almost starting to slip) on the same cart. The cart and donkey are accelerating at 0.24[m/s2] to the left without slippage.

Find-
a) The force by the donkey on the cart
b)The force of static friction by the surface on the donkey.

Newton's 2nd law

## The Attempt at a Solution

My first step is always to draw a FBD.

first part of that problem, I got an answer of 2065[N].

I'm having an issue with the second part.

I drew a FBD for the donkey and came up with for the x-axis-

Ff(static) - Fc/d = m(-a) and I believe that gets me the right answer(or at least a common answer among classmates).
You have a signage error here.
I just don't know why. Given this, the Ff(static) would be larger than Fc/d, so how exactly is the donkey moving? Thanks!

My only thought is that the Ffs is a force of the Earth on the donkey?
Yes , correct. When drawing FBD's, you always look at the forces acting on the object you have isolated. The donkey pushes back on the ground, so the ground friction pushes forward on the donkey. That is what makes it move.

So it would be...

-Ff(static) + Fc/d = m * -a

I just always thought the force of friction goes against movement... is that true only for kinetic(i.e. movement) friction?

noleguy33 said:
So it would be...

-Ff(static) + Fc/d = m * -a
yes, correct
I just always thought the force of friction goes against movement... is that true only for kinetic(i.e. movement) friction?
The force of friction acts against the relative movement (or impending movement) between the 2 surfaces. It applies to both static and kinetic friction, they can act either way depending on the relative movement between the 2 surfaces.
For the donkey, as in walking, where static friction applies, you must push back on the ground, so there is a tendency for your foot to slip backwards, so the friction force acts forward, opposite to the direction of the pending slip. In kinetic friction, a crate on a truck bed might start slipping backwards with respect to the truck bed as the truck accelerates forwards, so the kinetic friction acts forward on the crate, causing it to accelerate forward with respect to the ground.

As a scientist, it is important to approach problems like this with a clear understanding of the concepts and equations involved. In this case, you correctly used Newton's Second Law to set up equations for both the donkey and the cart.

In the first part of the problem, the cart is not accelerating, so the force of friction must be equal to the force applied by the donkey. This is why you were able to solve for the force of the donkey on the cart using the coefficient of kinetic friction.

In the second part, the cart and donkey are accelerating together, so the force of friction is no longer equal to the force applied by the donkey. Instead, the force of friction is now equal to the difference between the applied force and the force needed to accelerate the system. This is why you used the coefficient of static friction to solve for the force of friction in this case.

I can understand your confusion about the direction of the forces and how the donkey is moving. It is important to remember that the force of friction always acts in the opposite direction of motion, so in this case, the force of friction is acting to the right while the donkey and cart are moving to the left. This is why the force of friction is able to accelerate the system in the opposite direction.

In terms of the specific values and calculations, it is difficult to comment without seeing the exact numbers and equations used. However, it is important to double check your calculations and make sure they are consistent with the equations and concepts being applied. If you are still unsure, it may be helpful to consult with a classmate or a teacher for clarification.

## What is the definition of force of static friction in an FBD?

The force of static friction in an FBD (Free Body Diagram) is the force that opposes the movement of an object when it is in a state of rest or motionlessness. It is a type of contact force that acts between two surfaces in contact with each other.

## How is the force of static friction calculated in an FBD?

The force of static friction can be calculated by multiplying the coefficient of static friction (μs) between the two surfaces in contact and the normal force (N) exerted by one surface on the other. The formula is Fs = μsN.

## What factors affect the force of static friction in an FBD?

There are three main factors that affect the force of static friction in an FBD: the coefficient of static friction (μs), the normal force (N), and the roughness of the surfaces in contact. A higher coefficient of static friction and normal force will result in a higher force of static friction, while a rougher surface will also increase the force of static friction.

## How does the direction of force of static friction change in an FBD?

The direction of force of static friction always opposes the direction of potential motion, meaning it acts in the opposite direction of the force that is trying to move the object. This direction can change if the applied force on the object changes or if the object rotates, but it will always oppose the potential motion of the object.

## What happens if the force of static friction is greater than the applied force in an FBD?

If the force of static friction is greater than the applied force, the object will remain in a state of rest or motionlessness. This is because the force of static friction is strong enough to prevent the object from moving. However, if the applied force is greater than the force of static friction, the object will begin to move in the direction of the applied force.

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