Let's see what your diagram looks like. Do you know how to resolve vectors into components so that the vectors can be summed vectorially?alijan kk said:Homework Statement
Three forces each of magnitude 1N act from one corner towards the other corner of a square there sum has a magnitude nearest to :
Homework Equations
The Attempt at a Solution
diagonal of a square is equal to √2 , the answer should be 3√2
yes I know vector resolution, but I want a hint on how to picturize this questionChestermiller said:Let's see what your diagram looks like. Do you know how to resolve vectors into components so that the vectors can be summed vectorially?
Try laying the square out so that the corner at which the forces are applied is the lower left corner, and this corner coincides with the origin of an x-y Cartesian coordinate system. What are the components of the three forces in the x direction and in the y direction?alijan kk said:yes I know vector resolution, but I want a hint on how to picturize this question
the net force(x) is in the -x direction 3/√2 and the Fnet(y) is also 3/√2 in -y directionChestermiller said:Try laying the square out so that the corner at which the forces are applied is the lower left corner, and this corner coincides with the origin of an x-y Cartesian coordinate system. What are the components of the three forces in the x direction and in the y direction?
The question sounds like it is multiple choise, where you have not shown the choises. The correct answer might be very simple. Otherwise, you may still need to make a diagram and show details of your work. Which corner is "the other corner"? Are all the forces pointing in the same direction?alijan kk said:the net force(x) is in the -x direction 3/√2 and the Fnet(y) is also 3/√2 in -y direction
you are right its a multiple choice questionsFactChecker said:The question sounds like it is multiple choise, where you have not shown the choises. The correct answer might be very simple. Otherwise, you may still need to make a diagram and show details of your work. Which corner is "the other corner"? Are all the forces pointing in the same direction?
PS. Why do you think that the lay-out of the square effects the total magnitude of the force?
That's not what I get.alijan kk said:the net force(x) is in the -x direction 3/√2 and the Fnet(y) is also 3/√2 in -y direction
First, the diagonal of a unit square is √2 in length, but that isn't relevant in this problem. If you draw the diagram as @Chestermiller suggested, all forces extend out from the lower left corner of the square. All three forces have a magnitude of 1, so the force toward the upper right corner of the square doesn't reach that corner point.alijan kk said:diagonal of a square is equal to √2 , the answer should be 3√2
No.alijan kk said:the net force(x) is in the -x direction 3/√2 and the Fnet(y) is also 3/√2 in -y direction
Nor do I.Chestermiller said:That's not what I get.
That is the situation that I first assumed. But the statement given is "from one corner towards the other corner of a square". If they all point in the same direction, the answer is simple and quite different.Mark44 said:To expand on what Chestermiller said in post #4, draw your diagram with the lower left corner of the square at the origin, and with the three vectors extending from the origin toward the three other corners of the square.
Yes. However, since a square has four corners, "other corner of a square" is unclear. Do the vectors all start from one point, and point to the opposite corner of the square?FactChecker said:But the statement given is "from one corner towards the other corner of a square". If they all point in the same direction, the answer is simple and quite different.
Or at least a clear description of the problem.FactChecker said:We need to see a diagram of the vectors, where they start, and where they end. Or tell us that they all start at the same corner and point toward different corners.
you are right i made that mistake, but why it is not simply the sum of vector magnitudes, 3 ?Mark44 said:First, the diagonal of a unit square is √2 in length, but that isn't relevant in this problem. If you draw the diagram as @Chestermiller suggested, all forces extend out from the lower left corner of the square. All three forces have a magnitude of 1, so the force toward the upper right corner of the square doesn't reach that corner point.
Second, you apparently have multiplied √2 by 3 to get your answer, but that is incorrect for two reasons -- the magnitude of each force is 1, not √2, and the resultant of a force is not simply the sum of the magnitudes.No.
Nor do I.
Have they not taught you this in your course?alijan kk said:you are right i made that mistake, but why it is not simply the sum of vector magnitudes, 3 ?
i am just confused that why the corner to corner is diagonal why not the nearest corner could be the corner, and if the force is like you said from left down corner and extends to right upward corner ,then why can't we just add three vectors and get the magnitude of 3.Chestermiller said:Have they not taught you this in your course?
We still don't have a clear description of the problem. Do the vectors all point in the same direction, or does each one point to a different corner of the square?alijan kk said:you are right i made that mistake
actually i also wanted to know if the problem is correct or wrong, it is created by over physics proffesorMark44 said:We still don't have a clear description of the problem. Do the vectors all point in the same direction, or does each one point to a different corner of the square?
We can't give useful help if we don't know what the problem is.
It depends on what the problem is.alijan kk said:actually i also wanted to know if the problem is correct or wrong, it is created by over physics proffesor
As I already mentioned, the square has three other corners. Are the vectors pointing to different corners or are they all pointing toward the opposite corner?alijan kk said:Three forces each of magnitude 1N act from one corner towards the other corner of a square
yes, we have learned vector addition by head to tail rule and by law of cosines.Chestermiller said:What have you learned in your course about how to add vectors having different magnitudes and directions to obtain their resultant?
This is the only description given and the answer is nearest to 2.4Mark44 said:It depends on what the problem is.
From the original post:
As I already mentioned, the square has three other corners. Are the vectors pointing to different corners or are they all pointing toward the opposite corner?
These questions have different answers.
What about resolving into components, and adding the components?alijan kk said:yes, we have learned vector addition by head to tail rule and by law of cosines.
3 is only correct if they all point in the same direction. any "zig-zagging" reduces the magnitude of the sum.alijan kk said:you are right i made that mistake, but why it is not simply the sum of vector magnitudes, 3 ?
The purpose of finding the magnitude of 3 forces on a square is to understand the overall effect of these forces on the square. It allows us to determine the net force acting on the square, as well as the direction of this net force.
Typically, we consider external forces acting on the square, such as gravity and applied forces. Internal forces, such as tension within the square, are usually not considered as they cancel out in the calculation of net force.
The formula for finding the magnitude of 3 forces on a square is: Fnet = √(Fx² + Fy²), where Fx and Fy are the horizontal and vertical component of the net force, respectively.
To determine the direction of the net force, we use the trigonometric concept of finding the angle of a right triangle. We can use the inverse tangent function (tan⁻¹) to find the angle, which represents the direction of the net force.
The units of measurement for the magnitude of 3 forces on a square are typically in Newtons (N) or pounds (lbs), depending on the unit system being used. It is important to ensure that all forces are in the same unit before calculating the magnitude.