Two unlike vertical forces of 10N and Q N act at apoint A and be respectively where AB is horizontal and the length of 3m . The resultant of the 2 forces is a force of P N which acts at a point of X on the line of BA reduced such that AX is 4.5m.find P and Q
To find the values of P and Q, we can use the principle of moments or the principle of vector addition. Let's use the principle of vector addition in this case.
Since AB is horizontal and the resultant force acts at a point X on the line BA, we can draw a triangle ABC to represent the forces. Point A represents the point where the 10N force is applied, point B represents the point where the QN force is applied, and point C represents the point where the resultant force acts.
We are given that AB has a length of 3m, and AX has a length of 4.5m. Using these values, we can calculate the length of XC.
Let's consider the triangle AXB. By applying the Pythagorean theorem, we can find the length of XB.
XB^2 = AB^2 - AX^2
XB^2 = 3^2 - 4.5^2
XB^2 = 9 - 20.25
XB^2 = -11.25
The square root of -11.25 is not a real number, which means there is no triangle with the given dimensions. Hence, there is an error in the given information or the problem statement.
Please recheck the values or provide any additional information to proceed with the calculation and find the values of P and Q.