Question Video: Finding the Force That Causes the Resultant to Pass through a Certain Point on a Square | Nagwa Question Video: Finding the Force That Causes the Resultant to Pass through a Certain Point on a Square | Nagwa

Question Video: Finding the Force That Causes the Resultant to Pass through a Certain Point on a Square Mathematics • Third Year of Secondary School

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𝐴𝐡𝐢𝐷 is a square having a side length of 6 cm. Given that the line of action of the resultant of the forces is passing through the point 𝐸, where 𝐸 ∈ line segment 𝐴𝐡 and 𝐴𝐸 = 4 cm, determine the magnitude of 𝐹.

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Video Transcript

𝐴𝐡𝐢𝐷 is a square having a side length of six centimeters. Given that the line of action of the resultant of the forces is passing through the point 𝐸, where 𝐸 is in the line segment 𝐴𝐡 and 𝐴𝐸 equals four centimeters, determine the magnitude of 𝐹.

In this question, the resultant of the forces is nonzero. And therefore, we cannot simply balance horizontal and vertical forces to find 𝐹. What we are told is that the line of action of the resultant force passes through the point 𝐸. If the line of action of a force passes through a point, then its perpendicular distance from that point is zero. And therefore, its moment about that point is also zero. We therefore need to balance the moments of all of the forces about the point 𝐸 and solve for 𝐹. Let’s take positive moments to be anticlockwise about the pivot point as per convention.

Cycling through the forces then, starting with this force of two newtons from the point 𝐴, the line of action of this force passes through the point 𝐸. And therefore, its perpendicular distance from 𝐸 is zero, and therefore its moment about 𝐸 is also equal to zero. For the force of four newtons about the point 𝐴, this moment will clearly have an anticlockwise turning force about 𝐸. And therefore, the moment will be positive. The force of four newtons from the point 𝐷 will also have an anticlockwise turning force about 𝐸. So this moment will also be positive. The force of two newtons from the point 𝐡 will have a clockwise turning force about 𝐸. And therefore, this will be negative.

Assuming that 𝐹 is positive in the direction indicated on the diagram, it will have an anticlockwise turning force about 𝐸. So this will also be positive. If it turns out that 𝐹 is in fact negative in this direction, then this will work out in the equations since it will have the moment of the same magnitude but opposite sign.

Now, we need to determine the perpendicular distance from the pivot point 𝐸 of the line of action of all of these forces. For the force of two newtons from the point 𝐴, we have already established that this distance is zero. For the force of four newtons from the point 𝐴, we are given that 𝐴𝐸 equals four centimeters. And given that 𝐴𝐷 is a horizontal line and 𝐴𝐡 is a vertical line, the perpendicular distance of the line of action of this force will be four centimeters as well.

For the force of four newtons from the point 𝐷, we are given that the side length of the square is six centimeters. Therefore, the perpendicular distance of the line of action of this force from the point 𝐸 will be six centimeters. For the force of two newtons from the point 𝐡, we are again given that 𝐴𝐸 is equal to four centimeters. And therefore, 𝐡 is equal to six minus four, which is two centimeters. So the perpendicular distance of the line of action of this force from the point 𝐸 will be two centimeters.

For the perpendicular distance of the line of action of 𝐹 from the point 𝐸, we need to find this length here. This gives us a right triangle here. And we have already established that the length of the hypotenuse 𝐡𝐸 is two centimeters. Since the line 𝐡𝐷 clearly bisects the square, this angle in here will be 45 degrees. The length of the opposite side will therefore be given by the length of the hypotenuse two centimeters multiplied by the sin of 45 degrees. And this is equal to root two. Therefore, the perpendicular distance of the line of action of the force 𝐹 from the point 𝐸 is equal to root two.

Now, we can set the sum of all of these moments about the point 𝐸 equal to zero and solve for 𝐹. This gives us two times zero plus four times four plus four times six plus two times negative two plus root two times 𝐹 is equal to zero. Simplifying, this gives us 36 plus root two 𝐹 is equal to zero. Subtracting 36 from both sides and dividing by root two gives us 𝐹 equals negative 36 over root two. Multiplying the numerator and denominator by root two gives us negative 36 root two over two, which gives us negative 18 root two. We don’t need to worry about the negative. This simply indicates that 𝐹 is in the opposite direction to what is depicted on the diagram. The magnitude of 𝐹 is equal to 18 root two newtons.

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