A resultant force with a magnitude of 20 newtons can be produced by two concurrent forces having magnitudes of(1) 5.0 N and 10N(2) 20N and 20N(3) 20N and 50N(4) 30N and 5.0N
The correct answer and explanation is :
To determine which pair of forces can produce a resultant force of 20 newtons, we need to apply the principle of vector addition. Two forces are concurrent if they act at the same point, and their resultant can be found by either the law of cosines or the Pythagorean theorem, depending on the angle between the forces. For simplicity, we will analyze the cases based on the angle between the forces. The relevant equations to consider are:
- When forces are parallel (0 degrees between them):
[
R = F_1 + F_2
]
where (R) is the resultant force and (F_1) and (F_2) are the magnitudes of the two forces. - When forces are perpendicular (90 degrees between them):
[
R = \sqrt{F_1^2 + F_2^2}
]
Let’s now analyze each case:
Case 1: Forces of 5.0 N and 10 N
If the angle between the two forces is 0 degrees, the resultant force is:
[
R = 5.0 + 10 = 15 \, \text{N}
]
If the forces are at 90 degrees to each other, the resultant is:
[
R = \sqrt{5.0^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} \approx 11.18 \, \text{N}
]
Neither of these results is 20 N, so this combination does not work.
Case 2: Forces of 20 N and 20 N
If the angle between the forces is 0 degrees, the resultant force is:
[
R = 20 + 20 = 40 \, \text{N}
]
If the forces are perpendicular:
[
R = \sqrt{20^2 + 20^2} = \sqrt{400 + 400} = \sqrt{800} \approx 28.28 \, \text{N}
]
Again, neither of these results is 20 N.
Case 3: Forces of 20 N and 50 N
If the angle between the forces is 0 degrees:
[
R = 20 + 50 = 70 \, \text{N}
]
If the forces are perpendicular:
[
R = \sqrt{20^2 + 50^2} = \sqrt{400 + 2500} = \sqrt{2900} \approx 53.85 \, \text{N}
]
Neither of these results is 20 N.
Case 4: Forces of 30 N and 5.0 N
If the angle between the forces is 0 degrees:
[
R = 30 + 5.0 = 35 \, \text{N}
]
If the forces are perpendicular:
[
R = \sqrt{30^2 + 5.0^2} = \sqrt{900 + 25} = \sqrt{925} \approx 30.41 \, \text{N}
]
Neither of these results is 20 N.
Conclusion:
None of the combinations given in the choices (1-4) will produce a resultant force of exactly 20 N. It seems there may be some misinterpretation of the problem or the information provided. However, the analysis shows that a pair of forces producing a resultant force of 20 N would require careful adjustment of angles and magnitudes, which is not possible with the pairs provided.
Here is a vector diagram illustrating the addition of the forces for the four cases you provided. It shows how different combinations of forces result in various magnitudes of resultant forces. As discussed, none of the combinations yield a resultant of exactly 20 N.