# Forces acting on a moving crane

• Thickmax
In summary, the conversation discusses the process of transforming a vector equation to an algebraic equation and solving for a variable. The final equation is simplified by multiplying both the numerator and denominator by the same value, which is a fundamental algebraic rule. The answer is then simplified further by rearranging the terms and grouping them together. The process may vary depending on the target form and requires knowledge of algebraic rules.
Thickmax
Homework Statement
Engineering course top up physics module
Relevant Equations
See below

I've checked my answer many times and do not understand what I am missing

You probably made an error with signs. Remember that ##F_{resistance} \propto - \vec{e_x}## while ##F_{driving} \propto \vec{e_x}##.

Delta2
1/(M*g*v)*(P-C*v^3) No idea how this is the case

Delta2
Thickmax said:
1/(M*g*v)*(P-C*v^3) No idea how this is the case

Since ##\vec{F}_{total}.\vec{e}_y = 0##, we have $$\vec{F}_{N} = mg \vec{e}_y$$ $$\vec{F}_{Friction} = - \mu mg \vec{e_x}$$ and $$\vec{F}_{drag} = - C v^2 \vec{e}_x$$ As you did for your answer you then take ##\vec{F}_{total}.\vec{e}_x = 0## adn resolve the equation for ##\mu##.

Thickmax
The crane is moving with constant horizontal velocity. Hence the vector sum of the horizontal forces is zero, that is
$$\vec{F}_{resistance}+\vec{F}_{driving}=\vec{0}$$

In order to transform the above vector equation to an algebraic equation , you have to be careful with signs and conventions. If we make the convention that the positive direction is towards right , then $$\vec{F}_{driving}=\frac{P}{v}\hat x$$ that is it is positive force, but $$\vec{F}_{resistance}=-Cv^2\hat x-\mu mg\hat x$$, that is it is negative (towards the left).
Thus the initial vector equation will transform to the algebraic equation
$$-Cv^2-\mu mg+\frac{P}{v}=0$$ which you can solve for ##\mu## and you will get what you saying at post #3.

Lnewqban and Thickmax
Delta2 said:
The crane is moving with constant horizontal velocity. Hence the vector sum of the horizontal forces is zero, that is
$$\vec{F}_{resistance}+\vec{F}_{driving}=\vec{0}$$

In order to transform the above vector equation to an algebraic equation , you have to be careful with signs and conventions. If we make the convention that the positive direction is towards right , then $$\vec{F}_{driving}=\frac{P}{v}\hat x$$ that is it is positive force, but $$\vec{F}_{resistance}=-Cv^2\hat x-\mu mg\hat x$$, that is it is negative (towards the left).
Thus the initial vector equation will transform to the algebraic equation
$$-Cv^2-\mu mg+\frac{P}{v}=0$$ which you can solve for ##\mu## and you will get what you saying at post #3.
Thanks Very much!

But I'm struggling to simplify the last equation still.

-(Cv^2)-(UMG)+(P/v)=0 = -(Cv^2)+(P/V)=UMG

U=(-(Cv^2)+(P/v))/(MG)

Am I missing something?

Thickmax said:
Am I missing something?

No you got it right, you just need one last algebraic step: multiply both the numerator and the denominator of the fraction by v and then you 'll get what you say at post #3.

Thickmax
Delta2 said:
No you got it right, you just need one last algebraic step: multiply both the numerator and the denominator of the fraction by v and then you 'll get what you say at post #3.
How do you know when to do that? Is this method called something that i can read up on?

Many thanks again

Delta2
Thickmax said:
How do you know when to do that? Is this method called something that i can read up on?
Ehm its fundamental algebraic rule: IF you multiply both the numerator and the denominator of a fraction with the same thing, the fraction remains the same. That is $$\frac{a}{b}=\frac{ac}{bc}$$

Thickmax
Delta2 said:
Ehm its fundamental algebraic rule: IF you multiply both the numerator and the denominator of a fraction with the same thing, the fraction remains the same. That is $$\frac{a}{b}=\frac{ac}{bc}$$
Oh I see nowBut how do you know when to do that? Is it just to tidy up the equation and have ‘one line’ as the numerator?

Thickmax said:
Oh I see nowBut how do you know when to do that? Is it just to tidy up the equation and have ‘one line’ as the numerator?
Yes you can view it like this.

Thickmax
Thickmax said:
Oh I see nowBut how do you know when to do that? Is it just to tidy up the equation and have ‘one line’ as the numerator?
Yes, it is usual to avoid nested fractions in the simplified form. But more generally, you cannot assume the target form is what you would consider the simplest such. You have to be prepared to check whether there is some way of manipulating the one into the other.

Thickmax and Delta2
Got it! thanks

But then why does the answer have (1/M*G*V)*(P-C*V^3)

Why can't it just be U=(-(CV^3)+(P))/(MGV) = (P-(CV^3))/(MGV)

Think this is a stupid question as I know the answer will be the same, but I don't understand how to know/why you simplify this far.

It doesn't help that I don't know all the rules either ;)

$$\frac{1}{A}B=\frac{B}{A}$$

Thickmax

## What is the definition of a force?

A force is a push or pull that can cause an object to accelerate or change its motion.

## What are the different types of forces acting on a moving crane?

The different types of forces acting on a moving crane include:

• Gravity: the force that pulls the crane and any objects attached to it towards the ground
• Tension: the force exerted by the crane's cables or ropes
• Friction: the force that opposes the movement of the crane and can cause it to slow down or stop
• Aerodynamic forces: the forces exerted by air on the crane as it moves through it

## How do these forces affect the stability of a moving crane?

The forces acting on a moving crane can affect its stability in different ways. For example, if the force of gravity is greater than the tension in the crane's cables, it can cause the crane to tip over. Friction can also play a role in stability, as it can either help to keep the crane in place or make it more difficult to move. Properly balancing and controlling these forces is crucial for maintaining the stability of a moving crane.

## What safety precautions should be taken when operating a crane?

When operating a crane, it is important to follow all safety protocols and guidelines. This includes ensuring that the crane is properly maintained and inspected, using the correct equipment and techniques for the job, and following proper load limits and weight distribution. It is also important to have a trained and experienced operator in control of the crane at all times.

## How can the forces acting on a moving crane be calculated and controlled?

The forces acting on a moving crane can be calculated using mathematical formulas and principles of physics. This information can then be used to determine the necessary counterweights, tension in cables, and other factors to maintain the stability and control of the crane. Additionally, advanced technologies such as sensors and computer systems can help monitor and regulate these forces in real-time to ensure safe and efficient operation of the crane.

• Introductory Physics Homework Help
Replies
12
Views
4K
• Introductory Physics Homework Help
Replies
29
Views
2K
• Introductory Physics Homework Help
Replies
37
Views
2K
• Mechanical Engineering
Replies
8
Views
802
• Introductory Physics Homework Help
Replies
10
Views
354
• Introductory Physics Homework Help
Replies
13
Views
5K
• Introductory Physics Homework Help
Replies
1
Views
937
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
8K
• Introductory Physics Homework Help
Replies
8
Views
267