System of forces in equilibrium

[vp_]

A force $F_1=i-3j-2k$ at the point $-2i+9j$, another force $F_2=2i+j-3k$ at the point $-i+yj-k$ and a third force $F_3$ are equivalent to zero. Find $y$ for this to be possible. Find $F_3$ and its line of action in this case

I infer that "equivalent to zero" means this system of forces is in equilibrium. I understand that for equilibrium, the resultant force on the system has to be zero, and the sum of moments must also be zero.

I understand how to calculate moments ($M=r \times F$), but fail to see how I can deduce $y$ without any information on where $F_3$ acts. I have $F_3=-3i+2j+5k$ by working out $F_1+F_2+F_3=0$ but am stuck beyond that.

Any help would be appreciated.

 

[anathema]

Hi there,

To find the value of y, we can use the principle of moments. Since the system is in equilibrium, the sum of moments about any point must be zero. Let's choose the origin as our point of reference.

The moment due to F_1 is given by M_1 = (-2i+9j) x (i-3j-2k) = -27k
The moment due to F_2 is given by M_2 = (-i+yj-k) x (2i+j-3k) = (-3+2y)j + (2+3y)k

Since the sum of moments is zero, we can equate the two moments and solve for y:
-27k = (-3+2y)j + (2+3y)k
Equating the k components, we get:
-27 = 2+3y
Solving for y, we get y = -9

Therefore, for the system to be in equilibrium, y must be equal to -9.

To find the third force F_3, we can use the fact that the sum of forces must also be zero. Therefore, we have:
F_1 + F_2 + F_3 = 0
Substituting the values of F_1 and F_2, we get:
(-2i+9j) + (2i+j-3k) + F_3 = 0
Simplifying, we get:
F_3 = 3k-10j

The line of action of F_3 can be found by finding the point where F_3 intersects the xy-plane. Since the z component of F_3 is zero, we can ignore it and consider only the x and y components. Therefore, the line of action of F_3 lies in the xy-plane and passes through the point (0,-10,0).

I hope this helps! Let me know if you have any further questions.

 

[thomate1]

To find y, we can use the fact that the sum of moments must be zero. Since we know the position vectors and forces for F_1 and F_2, we can calculate their individual moments about the origin. We can then set up an equation using the fact that the sum of moments must be zero and solve for y.

M_1 = (-2i+9j) x (i-3j-2k) = -6i-2k
M_2 = (-i+yj-k) x (2i+j-3k) = 3j+2k
M_total = M_1 + M_2 + M_3 = 0
M_3 = -M_1 - M_2 = 6i+2k-3j-2k = 6i-3j
M_total = (6i-3j) + 3j+2k = 6i-2k = 0

Now we can set up an equation using the y component of the moment:

6i-2k = 0
yj = 0
y = 0

Therefore, for this system of forces to be in equilibrium, the point where F_2 is applied (yj) must be at the origin.

To find F_3, we can use the fact that the resultant force on the system must be zero. We can set up an equation using the x, y, and z components of the forces:

F_total = F_1 + F_2 + F_3 = 0
F_3 = -F_1 - F_2 = -(-3i+2j+5k) = 3i-2j-5k

The line of action for F_3 can be found by taking the cross product of the position vector and the force vector for F_3:

r x F_3 = (-i+yj-k) x (3i-2j-5k) = 5i+3j+2k

Therefore, the line of action for F_3 is in the direction of 5i+3j+2k.