Three forces act on an object. Two of the forces are at an angle of 105 degrees to each other and have magnitudes 23 N and 13 N. The third is perpendicular to the plane of these two forces and has magnitude 2 N. Calculate the magnitude of the force (in N) that would exactly counterbalance these three forces. (Round your answer to one decimal place.)
The correct answer and explanation is:
To calculate the magnitude of the force that exactly counterbalances the three given forces, we need to determine the resultant of the original three forces and then find a force equal in magnitude but opposite in direction.
Given:
- Force A = 23 N
- Force B = 13 N
- Angle between Force A and B = 105°
- Force C = 2 N (perpendicular to the plane of A and B)
We solve this in two steps:
- Combine Force A and Force B (coplanar forces using the law of cosines).
- Combine the result of A+B with Force C (perpendicular force, apply Pythagoras).
Step 1: Combine A and B using the Law of Cosines
Let RABR_{AB} be the resultant of A and B. RAB=A2+B2+2ABcos(θ)R_{AB} = \sqrt{A^2 + B^2 + 2AB\cos(\theta)}
Substitute values: RAB=232+132+2⋅23⋅13⋅cos(105∘)R_{AB} = \sqrt{23^2 + 13^2 + 2 \cdot 23 \cdot 13 \cdot \cos(105^\circ)} RAB=529+169+2⋅23⋅13⋅(−0.2588)R_{AB} = \sqrt{529 + 169 + 2 \cdot 23 \cdot 13 \cdot (-0.2588)} RAB=698−154.4728=543.5272≈23.3 NR_{AB} = \sqrt{698 – 154.4728} = \sqrt{543.5272} \approx 23.3\ \text{N}
Step 2: Combine RABR_{AB} with Force C (perpendicular)
Since Force C (2 N) is perpendicular to the plane of the other two, we can treat this as a 3D vector problem. The total resultant RR is the vector sum of RABR_{AB} and Force C, using the Pythagorean theorem in 3D: R=RAB2+C2=(23.3)2+(2)2R = \sqrt{R_{AB}^2 + C^2} = \sqrt{(23.3)^2 + (2)^2} R=542.89+4=546.89≈23.4 NR = \sqrt{542.89 + 4} = \sqrt{546.89} \approx 23.4\ \text{N}
Final Answer:
23.4 N\boxed{23.4\ \text{N}}
Explanation (300 words):
This problem involves the vector addition of three forces acting on an object. Two of these forces (23 N and 13 N) lie in the same plane and form an angle of 105° between them. The third force (2 N) acts perpendicular to this plane, making the problem three-dimensional.
To solve it, we first reduce the coplanar problem (the 23 N and 13 N forces) using the law of cosines, which allows us to calculate the magnitude of the resultant when two vectors form an angle. Since the cosine of 105° is negative (−0.2588), this indicates the two forces are pushing against each other to some degree, reducing the resultant force slightly. The computed resultant, approximately 23.3 N, represents the net force in the plane.
Then, we introduce the third force, which is perpendicular to the plane of the first two. When two vectors are perpendicular, their magnitudes combine using the Pythagorean theorem. Think of this as forming a right-angled triangle in three-dimensional space, where the hypotenuse is the overall resultant. By squaring the planar resultant (23.3 N) and adding the square of the perpendicular force (2 N), we find the total net force acting on the object.
To counterbalance all three forces, a single force equal in magnitude but opposite in direction must be applied. This results in a net force of zero, achieving equilibrium. The magnitude of this counterbalancing force is therefore 23.4 N, rounded to one decimal place.