To find the ratio of forces around a right triangle

In summary, the problem involves an ABC right-angled triangle with AB = 4a and BC = 3a, where forces P, Q, and R act along the directed sides. The resultant of these forces is a couple, and the goal is to find the ratios P:Q:R. The book answer is 4:3:5, but the correct answer is actually 15:20:12. This was determined by equating moments about A, B, and C, and finding that 20Q = 12R = 15P. A mistake in the calculations was pointed out and corrected.
  • #1
gnits
137
46
Homework Statement
To find the ratio of forces around a right triangle
Relevant Equations
Equate moments of couples
Could I please ask for help with the following:

ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively. Find the ratios P:Q:R if their resultant is a couple.

Book answer is 4 : 3 : 5

Here's my diagram:

triangle.png


Here's my working:

tan ACB = 4/3 and so the cos of ACB is 3/5 and the sin of ACB is 4/5

As we are given that the resultant is a couple we can equate moments about A, B and C respectively, giving:

4aQ = 4R/5 * 3a = 3aP

Which gives:

20Q = 12R = 15P

so P : Q : R = 15 : 20 : 12

Not the book answer,
thanks for pointing to my error,

Mitch.
 
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  • #2
gnits said:
20Q = 12R = 15P

so P : Q : R = 15 : 20 : 12
Say P=15. Then Q=20. But 20Q = 400 and 15P=225, so ##20Q \ne 15P##.
 
  • #3
Thanks, I see my error.

Mitch.
 

1. What is the formula for finding the ratio of forces around a right triangle?

The formula for finding the ratio of forces around a right triangle is the Pythagorean theorem, which states that the square of the length of the hypotenuse (the longest side of the triangle) is equal to the sum of the squares of the lengths of the other two sides. This can be written as c² = a² + b², where c is the length of the hypotenuse and a and b are the lengths of the other two sides.

2. How do you determine the forces in each direction of a right triangle?

To determine the forces in each direction of a right triangle, you can use trigonometric ratios. The sine, cosine, and tangent ratios can be used to find the forces in the vertical and horizontal directions. For example, if you know the length of the hypotenuse and the angle of one of the sides, you can use the sine ratio to find the force in the vertical direction.

3. Can the ratio of forces be greater than 1 in a right triangle?

No, the ratio of forces in a right triangle cannot be greater than 1. This is because the Pythagorean theorem states that the sum of the squares of the lengths of the other two sides must be equal to the square of the length of the hypotenuse. This means that the hypotenuse, which represents the largest force, must always be greater than or equal to the other two forces.

4. How is the ratio of forces used in real-world applications?

The ratio of forces is used in many real-world applications, such as engineering and physics. For example, in structural engineering, the ratio of forces can be used to determine the stability and strength of a building or bridge. In physics, the ratio of forces is used to calculate the acceleration of an object in motion.

5. What are some common mistakes when finding the ratio of forces in a right triangle?

One common mistake when finding the ratio of forces in a right triangle is using the wrong units. It is important to make sure that all measurements are in the same units before calculating the ratio. Another mistake is using the wrong side lengths in the Pythagorean theorem, which can lead to incorrect results. It is also important to pay attention to the direction of the forces when using trigonometric ratios, as using the wrong direction can result in an incorrect ratio.

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