A body is in equilibrium under the action of three force..one force is 4.0N acting due east and one is 5.0N in direction,
60°NE.calculate the magnitude of the third force.
You have two sides and the included angle. What is the law of cosines?
Draw your vector diagram, then use the law of cosines.
To calculate the magnitude of the third force, we can use the concept of vector addition.
First, let's break down the given forces into components. The first force of 4.0N acting due east has no north component (because it is acting due east) and only an east component of 4.0N.
The second force of 5.0N in the direction 60°NE can be split into two components: the east component and the north component. To find these components, we can use trigonometry.
The direction 60°NE can be broken down into two right-angled triangles: a 30° angle from the positive x-axis and a 60° angle from the positive y-axis.
Using trigonometry, we can calculate the east and north components of the force:
East component = force magnitude * cosine(angle) = 5.0N * cos(30°)
North component = force magnitude * sine(angle) = 5.0N * sin(30°)
Now, we can find the total east component and the total north component by summing up the respective components of the forces.
Total east component = 4.0N + 5.0N * cos(30°)
Total north component = 5.0N * sin(30°)
Since the body is in equilibrium, the total east component and the total north component should cancel out. Therefore, we can write the equation:
Total east component = -Total north component
Now, we can solve for the magnitude of the third force:
Total east component = -Total north component
4.0N + 5.0N * cos(30°) = -5.0N * sin(30°)
Solving this equation will give us the magnitude of the third force.