Resolving Complex Problems: Seeking Help for Unknown Forces and Solutions

[future_eng]

in the first problem i canot know how i can i get resultant force if i havenot the angles of force , i think a lot of this problem but i canot get its soluntion so please help me in resolving it .

i try also to resolve the second problem but i canot so please help me in resolving it also

 

[matiasmorant]

In the first excercise, you don't have the angles, but notice that, for example, the 40 force has a component 3a (the side of the square) in the y direction and a component of 3a/2 in the x direction (it reaches half the upper side). so, if it's angle with respect to the x-axis is O, then Tan(O)=(3a)/(3a/2)=2 so O=Arctan(2), about 63 degrees. similar with the others.

 

[y.moghadamnia]

try using the torques in the second one

 

[future_eng]

In the first excercise, you don't have the angles, but notice that, for example, the 40 force has a component 3a (the side of the square) in the y direction and a component of 3a/2 in the x direction (it reaches half the upper side). so, if it's angle with respect to the x-axis is O, then Tan(O)=(3a)/(3a/2)=2 so O=Arctan(2), about 63 degrees. similar with the others.

Thnx a lot matiasmorant , you helped me so much in resolving the first one . Really thanx

 

[future_eng]

try using the torques in the second one

Thanx for ur reply . i tried torques but it didnot help

 

[future_eng]

i resolved the first one but now please help me in resolving the second one .

 

[zgozvrm]

notice that, for example, the 40 force has a component 3a (the side of the square) in the y direction and a component of 3a/2 in the x direction

Uh... how did you know that the x-component of the 40 force is (3a)/2? Just because it looks like it doesn't make it so!

 

[zgozvrm]

In order for the 30 unit vector to extend from the left side to the lower right corner, it must be greater than 3a units long. Also, for the 40 unit vector to extend from the lower left corner to the top side, it must be less than the length of the diagonal, that is, less than 3a[itex]\sqrt2[/tex] units long. However, there is not enough information given to find the exact numerical value for the resultant vector; it can only be shown in terms of "a."

 

[zgozvrm]

However, there is not enough information given to find the exact numerical value for the resultant vector; it can only be shown in terms of "a."

Even then, we can only give a range of answers.
That is, we don't have enough information to give us either the vertical component of the 30 unit vector, or the horizontal component of the 40 unit vector.

Since we know that in order for a vector to extend from one side of a square to one of the square's vertices, it must be both longer than one of the square's sides AND shorter than the square's diagonal (as stated in my last post).


Simplifying, we know that
$$\frac{20}{3}\sqrt{2} < a < 10$$

or

9.428 < a < 10


This means that the vertical component of the 40 unit vector, V1, is between 28.284 and 30 (3a). Or 28.284 < V1 < 30. That gives us a horizontal component, H1, such that 26.458 < H1 < 28.284.

Likewise, the horizontal component of the 30 unit vector, H2, is 28.284 < H2 < 30 and the vertical component of the 30 unit vector, V2, is 0 < V2 < 10.


Finally, the vertical component of the vector resulting from the sum of the 30 and 40 unit vectors, V3, is 28.284 < V3 < 40 and the horizontal component, H3, is 54.742 < H3 < 58.284