Two linebackers simultaneously hit a quarterback: Force Diagram

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SUMMARY

The discussion centers on calculating the net force and acceleration of a quarterback being hit by two linebackers. The first linebacker applies a force of 500 N at 25° North of East, while the second linebacker applies a force of 450 N at 60° South of East. The net force acting on the quarterback, who has a mass of 90 kg, is determined to be +678.2 N in the x-direction and -178.4 N in the y-direction, resulting in an overall acceleration of +7.54 m/s² in the x-direction and -1.98 m/s² in the y-direction. Participants emphasize the importance of correctly interpreting the signs of the forces and converting the final results into compass directions.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Proficiency in trigonometric functions (sine and cosine)
  • Ability to draw and interpret force diagrams
  • Knowledge of vector addition and resultant forces
NEXT STEPS
  • Learn how to draw and interpret force diagrams accurately
  • Study vector addition techniques for forces at angles
  • Explore the concept of resultant forces using the Pythagorean theorem
  • Understand how to express forces and accelerations in terms of compass directions
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Students studying physics, particularly those focusing on mechanics and force analysis, as well as educators seeking to enhance their teaching methods in vector resolution and force diagrams.

Alice_in_Wonderland
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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!
 
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Alice_in_Wonderland said:
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.
 
haruspex said:
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?
 
Alice_in_Wonderland said:
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:
Alice_in_Wonderland said:
450 N, 60° S of E
Alice_in_Wonderland said:
Fbx=225N, Fby=389.7N
You have a sign wrong.
 
haruspex said:
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?
 
Alice_in_Wonderland said:
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
Alice_in_Wonderland said:
60° S of E
 
haruspex said:
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?
 
haruspex said:
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:)
 
Alice_in_Wonderland said:
I think it's only Fby that is a negative number because like you said, I'm considering E as positive.
Yes.
Alice_in_Wonderland said:
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.)
 
  • #10
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?
 
  • #11
Alice_in_Wonderland said:
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.
 
  • #12
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?
 
  • #13
haruspex said:
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..
 
Last edited:
  • #14
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."
 
  • #15
Physics 20 AB Dion said:
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..
 
  • #16
Alice_in_Wonderland said:
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.
Alice_in_Wonderland said:
For example, Fnetx= 678.2 N of E and Fnety= 178.4 S of E?
No, it will be just Fnet, not x and y. Do you know how to find the resultant of two forces at right angles? Hint: Pythagoras.
 

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