Force Vector Problem: Find Magnitude & Direction of F3

AI Thread Summary
The discussion revolves around solving a force vector problem involving three forces, where the goal is to find the magnitude and direction of the third force, F3, ensuring the vector sum equals zero. The user initially misapplies the equations for force components, leading to incorrect calculations for F3. Key corrections include using the correct equations for the sum of forces in both the x and y directions and properly calculating the components of F1 without cosine. Additionally, the direction of F3 should be specified as a positive angle relative to the negative x-axis. The conversation emphasizes the importance of correctly summing vector components and applying the Pythagorean theorem for accurate results.
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Homework Statement



Three forces are applied to an object, as shown in the figure. Force F1 has a magnitude of 26.8 Newtons (N) and is directed 30.0° to the left of the +y axis. Force F2 has a magnitude of 15.2 N and points along the +x axis. A third force F3 is applied such that the vector sum of the three forces is 0 N. What are (a) the magnitude and (b) direction of this third force? Specify the direction as a positive angle relative to the negative x axis, as shown in the figure.

Picture: http://s236.photobucket.com/albums/ff132/airpenny123/?action=view&current=ch01p_46.gif

Homework Equations



F3X = F1X + F2X
F3Y = F1Y + F2Y

F3= √F3X^2 + F3Y^2


The Attempt at a Solution



F1X = (-26.8N)(.8660)
F1X = -23.21N

F2X = 15.2N

F3X = -23.21N + 15.2N = -8.01N

F1Y = (26.8N)(0.5)
F1Y = 13.4N

F2Y = 0

F3Y = 13.4N + 0 = 13.4N

I found the components of F3X = -8.01 and F3Y = 13.4. When I plug the components into: F3= √F3X^2 + F3Y^2, I get 15.6115 N, which apparently is wrong. I've been on this problem forever and I don't know what I'm doing wrong.

As for this part of the question: "Specify the direction as a positive angle relative to the negative x axis, as shown in the figure." I am completely lost...

Thanks for any help.
 
Last edited:
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Welcome to PF :smile:

I see two problems.

Since the vector sum is zero, the equations should read

F1x + F2x + F3x = 0​

and similarly for the y-components.

Also, calculating the x-component of F1 does not involve cos(30).

As for this part of the question: "Specify the direction as a positive angle relative to the negative x axis, as shown in the figure." I am completely lost...

They just mean the angle from the -x direction.
 
why is that my prof taught us to get first the force sumation in X and in Y and use phytagorean theorem and then get the direction ..
 
Welcome to Physics Forums.
dgiordan said:
why is that my prof taught us to get first the force sumation in X and in Y and use phytagorean theorem and then get the direction ..
That is pretty much how you do this sort of problem, or anything that involves adding vectors. I'm not sure what specifically is confusing you about it?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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