Question Video: Finding the Resultant of Four Forces Acting on a Square at a Single Point | Nagwa Question Video: Finding the Resultant of Four Forces Acting on a Square at a Single Point | Nagwa

# Question Video: Finding the Resultant of Four Forces Acting on a Square at a Single Point Mathematics • Second Year of Secondary School

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The diagram shows a square, π΄π΅πΆπ·, of side 8 cm. The point πΈ is on the line segment π΅πΆ, where π΅πΈ = 6 cm. Forces of magnitudes 8 N, 20 N, 16β2 N, and 12 N act on π΄ as indicated on the diagram. Find the magnitude of their resultant.

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### Video Transcript

The diagram shows a square, π΄π΅πΆπ·, of side eight centimeters. The point πΈ is on the line segment π΅πΆ, where π΅πΈ is equal to six centimeters. Forces of magnitudes eight newtons, 20 newtons, 16 root two newtons, and 12 newtons act on π΄ as indicated on the diagram. Find the magnitude of their resultant.

We are told that the diagram shows a square with side length eight centimeters. Weβre also told that π΅πΈ is equal to six centimeters. Using our knowledge of the Pythagorean triples or the Pythagorean theorem, we know that π΄πΈ is of length 10 centimeters. We are asked to find the magnitude of the resultant force. In order to do this, we will begin by finding the sum of the forces in the π₯- and π¦-direction. We will take the positive π₯-direction to be the direction π΄π΅ and the positive π¦-direction to be in the direction π΄π·. Both the 20-newton and 16-root-two-newton forces do not act in either the horizontal or vertical direction. We will therefore need to calculate the horizontal and vertical components of these forces.

We will begin by letting the angle between the 20-newton force and the horizontal be π. Using our knowledge of right-angle trigonometry, we can begin by labeling the opposite, adjacent, and hypotenuse. As the sin of angle π is equal to the opposite over the hypotenuse, this is equal to six over 10, which simplifies to three-fifths. The cos of angle π is equal to the adjacent over the hypotenuse. Therefore, this is equal to four-fifths.

We will now let the horizontal component of the 20-newton force be π₯ and the vertical component be π¦. This means that π₯ is equal to 20 multiplied by cos π and π¦ is equal to 20 multiplied by sin π. The horizontal component is therefore equal to 16 newtons and the vertical component is equal to 12 newtons. We now know that there are two positive forces in the π₯-direction of eight newtons and 16 newtons. In the π¦-direction, we have positive forces of 12 newtons and 12 newtons.

The only force we have left to consider from the original diagram is that of 16 root two newtons. This is acting along the diagonal of our square in the direction πΆ to π΄. We know that the diagonal of a square makes an angle of 45 degrees with each of its sides. Using our knowledge of right-angle trigonometry once again, we have a horizontal component acting to the left equal to 16 root two multiplied by the cos of 45 degrees. The cos of 45 degrees is equal to root two over two. This means that the magnitude of the horizontal component is equal to 16 newtons. As this is acting in the negative π₯-direction, we can add negative 16 to our expression for π sub π₯. π sub π₯ is equal to eight plus 16 minus 16, which is equal to eight newtons.

The vertical component of our 16-root-two-newton force is acting downwards. This is equal to 16 root two multiplied by the sin of 45 degrees. The sin of 45 degrees is also equal to root two over two, which means that the magnitude of the vertical component is also equal to 16 newtons. Once again, this is acting in the negative direction, so π sub π¦ is equal to 12 plus 12 minus 16. 24 minus 16 equals eight. So, π sub π¦ is equal to eight newtons. Using our knowledge of the Pythagorean theorem, we know that the magnitude of the resultant π is equal to the square root of π sub π₯ squared plus π sub π¦ squared. In this question, π is equal to the square root of eight squared plus eight squared. This is the square root of 128, which simplifies to eight root two.

The magnitude of the resultant force is eight root two newtons.

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