Three forces act on a moving object

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SUMMARY

The discussion centers on calculating the third force required to maintain constant velocity on an object subjected to two forces: 80.0 N due north and 60.0 N due west. The net force was determined to be 100 N, with a direction of 53.13 degrees North of West. To counteract the existing forces, the third force must have a magnitude of 100 N directed 53.13 degrees South of East. This ensures that the object continues to move with a constant velocity.

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  • Understanding of vector addition and force equilibrium
  • Familiarity with Cartesian coordinate systems
  • Knowledge of trigonometric functions, specifically tangent and inverse tangent
  • Basic principles of Newton's laws of motion
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of force equilibrium and vector resolution in mechanics.

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Homework Statement



Three forces act on a moving object. One force has a magnitude of 80.0 N and is directed due north. Another force has a magnitude of 60.0 N and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

Force 1 = 80.0 N due north.
Force 2 = 60.0 N due west.

Homework Equations



a^2 + b^2 = c^2
Theta = tan^-1(fy/fx)

The Attempt at a Solution



First I started by plotting the problem on a cartesian plane. I then took my known quantities of 80.0 N due north (along the positive y axis) and 60.0 N due west (along the negative x axis) and used pythag to find the net force, the answer of which was 100 N.

I then attempted to find the direction using Theta = tan^-1(fy/fx) = tan^-1(80/60) = 53.13.

My question is, am I right in assuming that I must counteract the first two forces with the third to ensure a constant velocity? If so I would appreciate any help in this regard.
 
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Careful with your direction. 53.13 degrees with respect to what? West of North? North of West? Otherwise, your assumption is correct.
 
lewando said:
Careful with your direction. 53.13 degrees with respect to what? West of North? North of West? Otherwise, your assumption is correct.

Thanks. I will be sure to always define that in future :smile:

So the next step in my problem having confirmed my original assumption (thanks again) would be to determine the magnitude of the third force and it's direction:

The net force I worked out to be 100 N and the direction to be 53.13 degrees North of West :smile:

So the magnitude would have to be 100 N in the opposite direction being 53.13 degrees South of East?

I appreciate all help as I have just begun studying again and I want to fully grasp each concept as I go along.

Thanks.
 
So the magnitude would have to be 100 N in the opposite direction being 53.13 degrees South of East?

Yep.:approve:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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