Question

Four forces acting at a point. The first is 200N acting due north,the second of 100N acting due south, the third of 500N acting due eastand the fourth of 300N acting due west. What is the magnitude and direction of the resultant force?

Answer 1

Answer:

Explanation: please see the attachment....

Answer 2

Concept:

• Vector addition
• Calculating the magnitude of a resultant vector
• Determining the direction of a resultant vector

Given:

• Let north be the positive x-axis, south be the negative x-axis
• Let east be the positive y-axis, west be the negative y-axis
• Force acting northwards F1 = 200 N, F1 = 200 i
• Force acting south F2 = 100 N, F2 = -100 i
• Force acting east F3 = 500N, F3 = 500 j
• Force acting west F4 = 300N, F4 = -300 j

Find:

• The magnitude of the resultant force
• The direction of the resultant force

Solution:

Net force acting along the x-axis = F1+F2 = 200 i + (-100 i)

The net force acting along the x-axis = 100 i

Net force acting along the y-axis = F3+F4 = 500 j + (-300 j)

The net force acting along the y-axis = 200 j

The resultant net force = √100²+200²

The resultant net force = √50000

The resultant net force = 223.6 N

To determine the direction of the resultant force

tanФ = y/x = 200/100 = 2

Ф= tan⁻¹ (2)

Ф= 63.4° to the horizontal

The resultant force has a magnitude of 223.6 N and the direction is 63.4° to the east.

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