A pendulum bob of weight – W [N] is held at an angle θ from the vertical horizontal force F [N] is shown.

A pendulum bob of weight – W [N] is held at an angle θ from the vertical horizontal force F [N] is shown. The tension [N] is given by the equation: T = W * cos(θ)

The Correct Answer and Explanation is:

The correct answer is C. √5 N.

Explanation:

This problem involves a system in static equilibrium. According to Newton’s First Law, for an object to be held stationary, the net force acting on it must be zero. This means the vector sum of all forces must equal zero. In this case, there are three forces acting on the pendulum bob:

1. Weight (w): The force of gravity, given as 1 N, acting vertically downwards.
2. Horizontal Force (F): An external force, given as 2 N, acting horizontally to the right.
3. Tension (T): The force exerted by the string, acting upwards along the string at an angle θ from the vertical. We need to find the magnitude of this tension.

Since the net force is zero, we can solve this problem in two ways: by resolving forces into components or by using vector addition graphically.

Method 1: Resolving Forces into Components

We can break down the forces into their horizontal (x) and vertical (y) components. For the system to be in equilibrium, the sum of the forces in each direction must be zero.

• Vertical forces (y-direction): The upward vertical component of the tension must balance the downward force of weight.
  ΣFy = T_vertical – w = 0
  T * cos(θ) = w = 1 N
• Horizontal forces (x-direction): The leftward horizontal component of the tension must balance the rightward horizontal force F.
  ΣFx = F – T_horizontal = 0
  F = T * sin(θ) = 2 N

We now have two equations:

1. T * cos(θ) = 1
2. T * sin(θ) = 2

By squaring and adding these two equations, we can eliminate θ:
(T * cos(θ))² + (T * sin(θ))² = 1² + 2²
T²cos²(θ) + T²sin²(θ) = 1 + 4
T²(cos²(θ) + sin²(θ)) = 5

Using the trigonometric identity cos²(θ) + sin²(θ) = 1:
T²(1) = 5
T = √5 N

Method 2: Vector Addition (Geometric Approach)

Since the three forces sum to zero, if we arrange them head-to-tail, they will form a closed triangle. The weight vector (w) points down, and the applied force vector (F) points to the right. As these two forces are perpendicular, they form the legs of a right-angled triangle. The tension vector (T) must close this triangle, acting as the hypotenuse.

Using the Pythagorean theorem for this right-angled force triangle:
T² = w² + F²
T² = (1 N)² + (2 N)²
T² = 1 + 4 = 5
T = √5 N

Both methods confirm that the tension in the string is √5 N.