Question

If three forces acting in equilibrium at a point are 3 N, 4 N, and 5 N, the angle between the 3 N and 4 N forces is closest to:

• A. 36.87 degrees
• B. 5(3)13 degrees
• C. 90 degrees
• D. 120 degrees

Hint:

Equilibrium of Three Forces. If three forces are in equilibrium, they form a closed vector triangle. If their magnitudes form a Pythagorean triple (a, b, c where \(a^2+b^2=c^2\)), then the angle between the forces with magnitudes 'a' and 'b' must be 90 degrees.

Answer 1

The Correct Option is C

If three forces acting at a point are in equilibrium, their vector sum is zero.
This means the forces can be represented as the sides of a closed triangle when placed head-to-tail.
The magnitudes of the forces are 3 N, 4 N, and 5 N.
We recognize that 3, 4, and 5 form a Pythagorean triple, since \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\).
This implies that the force triangle formed by these vectors must be a right-angled triangle.
In equilibrium, the vector sum is zero: \(\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0\).
This means \(\vec{F}_1 + \vec{F}_2 = -\vec{F}_3\).
The resultant of the 3 N and 4 N forces must be equal in magnitude and opposite in direction to the 5 N force.
Since the magnitude of the resultant (\(R\)) of the 3 N and 4 N forces must be 5 N, and we know \(3^2 + 4^2 = 5^2\), the angle (\(\theta\)) between the 3 N and 4 N forces must satisfy the law of cosines for the resultant: $$ R^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos\theta $$ $$ 5^2 = 3^2 + 4^2 + 2(3)(4) \cos\theta $$ $$ 25 = 9 + 16 + 24 \cos\theta $$ $$ 25 = 25 + 24 \cos\theta $$ $$ 0 = 24 \cos\theta $$ $$ \cos\theta = 0 $$ $$ \theta = 90^\circ $$ The angle between the 3 N and 4 N forces is 90 degrees.