Question on force and simultaneous equations

[Kalix]

Homework Statement

Question: If an 80kg object experiences a normal force of 510N, and an unknown angled pull force. The coefficient of friction is 0.33. If the object moves with a constant velocity, what is the direction (angle) and magnitude (size) of the pull force?

m=80kg
a=0m/s
Fx=FcosΘ
Fy=FsinΘ
μ(mu)=0.33
Fk=510N x 0.33
g=-9.81
weight=80kg x 9.81

Homework Equations

Fx-fk=max (the "x" is a coefficient)
Fy-fk=may (the "y" is a coefficient)
Fs=μs x Fn
Fk=μk x Fn

The Attempt at a Solution

First I set up a simultaneous equation using Fx-fk=max and Fy-fk=may. This is what I got.
80kg+0m/s=FcosΘ -(510N)(0.33)
80kg+0m/s=FsinΘ -(80kg)(9.81)

This is where I get stuck. I am really bad at solving simultaneous equations.

 

[cepheid]

a=0m/s

Careful. Units are crucial. This should read a = 0 m/s² or a = 0 m/s^2 in plain text. To produce superscripts and subscripts , trying using the buttons labelled X² and X₂ above the reply box. Alternatively, you can manually put SUP and SUB tags around text. For example, the input [noparse]ma_x[/noparse] will produce the output ma_x

Homework Equations

Fx-fk=max (the "x" is a coefficient)
Fy-fk=may (the "y" is a coefficient)

I assume you mean that the x and y are subscripts (not coefficients). Again, see the note above about subscripts.

In any case, the second equation there doesn't make any sense, because there is no friction in the vertical direction. Friction always acts to oppose the sliding motion between the two surfaces that are in contact. In this case, assuming that the sliding is horizontally across the surface, then the friction will also act horizontally, in the opposite direction.

The Attempt at a Solution

First I set up a simultaneous equation using Fx-fk=max and Fy-fk=may. This is what I got.
80kg+0m/s=FcosΘ -(510N)(0.33)
80kg+0m/s=FsinΘ -(80kg)(9.81)

This is where I get stuck. I am really bad at solving simultaneous equations.

This looks better, but the plus signs in red should be multiplication signs. It's F_net = ma, not F_net = m+a.

 

[Fewmet]

The Attempt at a Solution

First I set up a simultaneous equation using Fx-fk=max and Fy-fk=may. This is what I got.
80kg+0m/s=FcosΘ -(510N)(0.33)
80kg+0m/s=FsinΘ -(80kg)(9.81)

This is where I get stuck. I am really bad at solving simultaneous equations.

You are correct to apply the second law of motion, but it says the sum of the forces equals the product of mass and acceleration. You've set the the net force equal to the sum of mass and acceleration. (Also note that the units of accelerations meters per second per second, which is usually written as m/s².)

So the sum of the forces in the forward direction is 80 kg * 0 m/s² = 0 N.

You could also look at the sum of the forces vertically, and that also equals zero (assuming the 80 kg mass moves horizontally).

No simultaneous questions are necessary. Does this let you see how to proceed?

 

[Kalix]

To fewmet: unfortunately my teacher requires us to do these problems with simultaneous equations. So if you do know how to do these types of equations that would be great.

To cepheid:I will be more careful with my signs/numbers. But could you still help me start the equation. I just don't know what the first step is.

 

[cepheid]

To fewmet: unfortunately my teacher requires us to do these problems with simultaneous equations. So if you do know how to do these types of equations that would be great.

To cepheid:I will be more careful with my signs/numbers. But could you still help me start the equation. I just don't know what the first step is.

You should have read the last sentence of my post more carefully. I told you that you ALREADY had the correct system of equations except for that error in which you added instead of multiplying (which I highlighted in red).

So, you HAVE the equations:

FcosΘ - μN = 0

FsinΘ - mg = 0

You have two equations and two unknowns, which means that you can arrive at a solution.

 

[Kalix]

I understand that I have the correct equations but I don't know how to solve them? As I said I have great difficulty in isolating variables so I guess that is what I am asking.

 

[Fewmet]

You should have read the last sentence of my post more carefully. I told you that you ALREADY had the correct system of equations except for that error in which you added instead of multiplying (which I highlighted in red).

So, you HAVE the equations:

FcosΘ - μN = 0

FsinΘ - mg = 0

You have two equations and two unknowns, which means that you can arrive at a solution.

I see I was hasty to assert there is no need for simultaneous equations. I should have looked at the problem on paper...

 

[cepheid]

I understand that I have the correct equations but I don't know how to solve them? As I said I have great difficulty in isolating variables so I guess that is what I am asking.

With 2 unknowns, the simplest method is:

- use the first equation to solve for one unknown in terms of the other one.

- substitute the expression you obtain into the second equation, which eliminates the first unknown altogether. Now use the equation to solve for the second unknown

 

[Fewmet]

I understand that I have the correct equations but I don't know how to solve them? As I said I have great difficulty in isolating variables so I guess that is what I am asking.

We really cannot do the algebra for you on homework. As cepheid said,

So, you HAVE the equations:

FcosΘ - μN = 0

FsinΘ - mg = 0

One general approach is to solve both for a variable that have in common (F or Θ, in this case, and you might find it easier to solve for F). Then set those two expressions equal to each other and solve for the other variable the original equations have in common (Θ or F).

(It might be helpful to recall from your trigonometry that sin Θ/cos Θ = tan Θ).