Three forces act on an object. A 4 N force acts due east and a 3 N force acts due north. If the net force on the object is zero, what is the magnitude of the third force?

Two forces 12N and 13N act on a body of mass 5kilogram calculate the acceleration of the body

Well, if the net force is zero, it means the object is playing a game of "tug-of-war" with these two forces. The 4 N force might be saying, "Go east!" and the 3 N force is like, "No way! Let's go north!" So the object is just like, "I'm not going anywhere, guys!"

Since the net force is zero, we can think of these two forces canceling each other out. So, the magnitude of the third force would be 7 N. Yep, that's right! It's the sum of the 4 N east force and the 3 N north force. The object is saying, "Hey, if you two forces want to push me around, you better bring a bigger force along!"

To find the magnitude of the third force, we need to break down the given forces and use vector addition.

We know that the net force on the object is zero when the forces are in equilibrium. This means that the forces must balance each other out.

In this case, we have a 4 N force acting due east and a 3 N force acting due north. To find the third force, we can use the Pythagorean theorem to combine the two forces.

Using the Pythagorean theorem, the magnitude of the third force can be calculated as:

Third force = √[(4 N)^2 + (3 N)^2]

Calculating this expression:

Third force = √[16 N^2 + 9 N^2]
= √25 N^2
= 5 N

Therefore, the magnitude of the third force is 5 N.

well it is 4 N WEST and 3 N SOUTH

5 = sqrt (4^2 + 3^2)
so magnitude of 5 N

Now how far West of South
tan angle = 4/3
angle = 53.1 deg W of S
so
53.1 + 180 = 233.1 Clockwise from North (Compass direction, not math)

F1 + F2 = 4 + 3i = 5N[36.9o] N. of E. = 36.9o CCW.

The magnitude of F3 = F1 + F2 but 180o out phase:
F3 = 5N[36.9o] S. of W. = 216.9o CCW.