Force of static friction in an FBD

[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.

Homework Equations

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?

 

[PhanthomJay]

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.

Homework Equations

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.

 

[noleguy33]

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?

 

[PhanthomJay]

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.

 

[rwisz]

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.