Two linebackers simultaneously hit a quarterback: Force Diagram

[Alice_in_Wonderland]

Homework Statement

Two linebackers simultaneously hit a quarterback, who has a mass of 90 kg. One linebacker exerts a force on the quarterback of 500 N, 25° N of E. The other linebacker exerts a force on the quarterback of 450 N, 60° S of E. Determine the net force acting on the quarterback and the quarterback’s acceleration.

I have to draw a force diagram that represents this question as well..

Given:
m= 90kg
Applied force linebacker 1 on quarterback (FaL1Q)= 500N
Applied force linebacker 2 on quarterback (FaL2Q)= 450N
Fg= -882.9N

Homework Equations

F= ma
Trigonometric functions (cosine and sine)

The Attempt at a Solution

I have no idea how to draw a force diagram of this question but I think I managed to find a solution although I don't know if it's right..

Using sine and cosine ratios,
Fax= 453.2N, Fay=211.3N
Fbx=225N, Fby=389.7N

Therefore, Fnet in the x is +678.2N (453.2N+225N)
and acceleration is +7.54m/s/s (687.2N/90kg)

Is this solution correct? Is this question asking for me to get net force in the x or y? Do both answers actually have positive signs to them?

*************COULD YOU PLEASE TELL ME HOW TO DRAW A FORCE DIAGRAM OF THIS?*********

Thank you in advance!

 

[haruspex]

Fax= 453.2N, Fay=211.3N
Fbx=225N, Fby=389.7N

Be careful with signs. I assume you are taking N and E as positive y and positive x.

 

[Alice_in_Wonderland]

Be careful with signs. I assume you are taking N and E as positive y and positive x.

Yes. Anything that is not with negative signs have positive sign to them. Could you please tell me if my attempt at a solution is correct or not?

 

[haruspex]

Yes. Anything that is not with negative signs have positive sign to them. Could you please tell me if my attempt at a solution is correct or not?

I am trying to draw your attention to this:

450 N, 60° S of E

Fbx=225N, Fby=389.7N

You have a sign wrong.

 

[Alice_in_Wonderland]

I am trying to draw your attention to this:You have a sign wrong.

Oh! so are you telling me that Fbx= -225N and Fby=-389.7N or only Fby=-389.7N?

 

[haruspex]

Oh! so are you telling me that Fbx= -225N and Fby=-389.7N or only Fby=-389.7N?

Which do you think? You have chosen that E should be positive, so W negative, and that N should be positive, so S negative. So what signs should you use for the components of

60° S of E

 

[Alice_in_Wonderland]

Which do you think? You have chosen that E should be positive, so W negative, and that N should be positive, so S negative. So what signs should you use for the components of

I think it's only Fby that is a negative number because like you said, I'm considering E as positive. Is that right? But in that case, my answer won't change, right?

 

[Alice_in_Wonderland]

Which do you think? You have chosen that E should be positive, so W negative, and that N should be positive, so S negative. So what signs should you use for the components of

Sorry for asking you a lot of questions.. I just want to make sure everything's correct:)

 

[haruspex]

I think it's only Fby that is a negative number because like you said, I'm considering E as positive.

Yes.

But in that case, my answer won't change, right?

Only because you failed to use the y forces.
Find the net y force. You then need to combine the net x and y forces to find the magnitude and direction of the overall force. (At least, I assume that is the form of answer they want. Alternatively, they might be happy if you keep the answers as separate x and y forces and accelerations.)

 

[Alice_in_Wonderland]

Oh, I see.. I thought I'm only getting Fnet in the x axis! Thank you.. Then is the answer for x force=+678.2N, acceleration=+7.54m/s/s & y force=-178.4N and acceleration=-1.98m/s/s?

 

[haruspex]

Oh, I see.. I thought I'm only getting Fnet in the x axis! Thank you.. Then is the answer for x force=+678.2N, acceleration=+7.54m/s/s & y force=-178.4N and acceleration=-1.98m/s/s?

Yes, but if you are going to leave them in component form then you should at least convert back to using compass directions, e.g. force = 678.2N E, 178.4N S.
And as I wrote, they probably want you to express each of force and acceleration as a single magnitude with a compass direction as some number of degrees S of E.

 

[Alice_in_Wonderland]

So then may I use the same compass directions as the question provides me with the final answer? For example, Fnetx= 678.2 N of E and Fnety= 178.4 S of E?

 

[Alice_in_Wonderland]

express each of force and acceleration as a single magnitude

How do I express force and acceleration as single magnitude? Don't they have different units? And how do I express with number of degrees? I've been looking for how to do that for about 20 mins but I'm not sure..

 

[Physics 20 AB Dion]

Hi there Alice, were you able to do question 2? If so please let me know asap. "Dylan is rushing to catch a flight. As he walks in the airport he pulls his suitcase behind him, which is rolling on its wheels. The mass of the suitcase is 20.0 kg. Dylan pulls on the handle so that he exerts a force on the handle of 100 N, 30° above the horizontal. Determine the normal force exerted by the floor on the suitcase."

 

[Alice_in_Wonderland]

Hi there Alice, were you able to do question 2? If so please let me know asap. "Dylan is rushing to catch a flight. As he walks in the airport he pulls his suitcase behind him, which is rolling on its wheels. The mass of the suitcase is 20.0 kg. Dylan pulls on the handle so that he exerts a force on the handle of 100 N, 30° above the horizontal. Determine the normal force exerted by the floor on the suitcase."

https://www.physicsforums.com/threads/determine-normal-force.959335/
In here there is my attempt at a solution but I'm not sure if that's the right answer..

 

[haruspex]

How do I express force and acceleration as single magnitude? Don't they have different units?

No, I wrote a single magnitude for each: one for force and one for acceleration.

For example, Fnetx= 678.2 N of E and Fnety= 178.4 S of E?

No, it will be just F_net, not x and y. Do you know how to find the resultant of two forces at right angles? Hint: Pythagoras.