Calculating Mass using F=ma and Friction

[Drew552]

Homework Statement

Well i have the force which equals 500N and I don't have a given acceleration so should I be substituting 9.81 as the acceleration? or is their another way to find it given that I also have the friction which equals 125N. So which way is the right way of going?

Homework Equations

F=Ma

The Attempt at a Solution

Well what I've done is 500=m x 9.81, which goes to 500/9.81=m, which then gives me 51.02kg, so is that correct because it doesn't seem right to me.

Any help would be much appreciated,

Thanks.

 

[Doc Al]

Can you please state the exact problem you are asked to solve, word for word.

 

[Drew552]

Its a fill in the chart question so its not really a word for word one, i have all the values but this one. If i can provide anything else please tell me.

 

[Doc Al]

You need to describe the situation a bit more. You have the force on what? What is it doing?

 

[Drew552]

Well its on a friction simulator online. I'm pushing on a unknown mass with 500N of applied force. The friction force comes out as 125N and the sum of the forces is 375N to the right direction. I am asked to find the mass and the weight of the object. I want to know the mass of the object which in turn will let me work out the weight with W=mg.

Thanks.

 

[Doc Al]

I'm pushing on a unknown mass with 500N of applied force.

OK. Across a horizontal surface?

The friction force comes out as 125N and the sum of the forces is 375N to the right direction.

So the only horizontal forces involved are the applied force and the friction?

I am asked to find the mass and the weight of the object.

Are you given any additional information? Such as the acceleration or the coefficient of friction?

 

[Drew552]

Yes it is being pushed on an horizontal surface. And yes the only horizontal forces are the force and friction. No i am not given the acceleration but I am assuming it could possibly be 9.81 and it doesn't give me the coefficient of friction.

 

[Doc Al]

No i am not given the acceleration but I am assuming it could possibly be 9.81

That's the acceleration due to gravity. No reason to think the mass has that acceleration.

I do not see that enough information has been given to solve the problem. Does the diagram or chart have any other info in it? Or a previous problem?

 

[Drew552]

Am i allowed to attach a picture of the chart and also the picture of the simulator??

 

[Doc Al]

Am i allowed to attach a picture of the chart and also the picture of the simulator??

Go ahead.

 

[Drew552]

.

 

[Drew552]

 

[Doc Al]

So you have plenty of other info! Hint: Use the data for the other masses to figure out the coefficient of friction. (In fact, if you understand the chart, you don't even need to calculate the coefficient. You can just read off the mass that you need.)

 

[Drew552]

I do? Well point taken then, thanks for the hint, ill try to to do my best at getting the answer but still unsure how to get the friction coefficient

 

[Doc Al]

Hint: What's the basic relationship for kinetic friction that uses the coefficient of friction?

 

[Drew552]

F(kinetic)= u(coefficient) x F(normal)??

 

[Doc Al]

F(kinetic)= u(coefficient) x F(normal)??

Exactly! And in these examples, what creates the normal force?

 

[Drew552]

a entity pushing against something thus making the kinetic energy?

 

[Drew552]

so if i had a 20N force pushing a against a box the normal force would equal 20N right?

 

[Doc Al]

The normal force is the force pushing each mass against the surface. What is that force?

 

[Drew552]

The force is the 2 entities rubbing together?

 

[Doc Al]

The force is the 2 entities rubbing together?

Sure, but what supplies that force? It's not the applied force, since that force is horizontal. Hint: What's the only vertical force acting on the masses?

 

[Drew552]

gravity?

 

[Doc Al]

gravity?

Of course!

Now pick anyone of the masses and use the given data to figure out the coefficient of friction (which is a constant for this problem).

 

[Drew552]

so it would be 40=u+9.8 Assuming 40 is the mass

 

[Drew552]

so is the coefficient of friction constant over one horizontal surface no matter the mass?

 

[Doc Al]

so is the coefficient of friction constant over one horizontal surface no matter the mass?

Yes, assume for this problem that the coefficient of friction is constant regardless of mass. (Meaning: Assume the masses are all the same material.)

Pick one of the masses that are given. Find the normal force, which is just the weight. You are given the friction force. Use those values to calculate the coefficient.

Once you have the coefficient, you can use it with the friction force to calculate the unknown mass.

 

[Drew552]

Ok, thank you for all your help, I think I've got it down now. ill post again if i come to any complications.

Thanks a lot :)