The angle between the two forces of magnitude 20 N and 15 N is 60°; the 20 N force being horizontal. Determine the resultant in magnitude and direction, if: (a) the forces are pulls; and (b) the 15 N force is a push and 20 N force is a pull

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Two Forces Problem Solution

Problem Statement

The angle between the two forces of magnitude 20 N and 15 N is 60°; the 20 N force being horizontal. Determine the resultant in magnitude and direction, if:

(a) the forces are pulls; and

(b) the 15 N force is a push and 20 N force is a pull.

Force diagram showing two forces at an angle

Given Data

First Force P = 20 N (horizontal)
Second Force Q = 15 N
Angle Between Forces θ = 60°

Solution: Case (a) – Both Forces are Pulls

Determining the Magnitude of the Resultant

Step 1: Using the law of cosines to find the magnitude of the resultant R:

$$R^2 = P^2 + Q^2 + 2PQ\cos\theta$$
\begin{align} R^2 &= (20\text{ N})^2 + (15\text{ N})^2 \\ &+ 2 \times 20\text{ N} \times 15\text{ N} \times \cos(60°) \end{align}
\begin{align} R^2 &= 400\text{ N}^2 + 225\text{ N}^2 \\ &+ 600\text{ N}^2 \times 0.5 \end{align}
$$R^2 = 400\text{ N}^2 + 225\text{ N}^2 + 300\text{ N}^2$$
$$R^2 = 925\text{ N}^2$$
$$R = \sqrt{925\text{ N}^2} = 30.4\text{ N}$$

Determining the Direction of the Resultant

Step 2: Using the tangent formula to find angle α (the angle the resultant makes with the 20 N force):

$$\tan\alpha = \frac{Q\sin\theta}{P + Q\cos\theta}$$
\begin{align} \tan\alpha &= \frac{15\text{ N} \times \sin(60°)}{20\text{ N} + 15\text{ N} \times \cos(60°)} \end{align}
\begin{align} \tan\alpha &= \frac{15\text{ N} \times 0.866}{20\text{ N} + 15\text{ N} \times 0.5} \end{align}
$$\tan\alpha = \frac{12.99\text{ N}}{20\text{ N} + 7.5\text{ N}}$$
$$\tan\alpha = \frac{12.99\text{ N}}{27.5\text{ N}} = 0.472$$
$$\alpha = \tan^{-1}(0.472) = 25.3° \approx 25°18’$$

Case (a) Results:

Magnitude of resultant force R = 30.4 N

Direction of resultant force = 25°18′ with the 20 N force

Solution: Case (b) – 15 N Force is a Push, 20 N Force is a Pull

Determining the Magnitude of the Resultant

Step 1: When one force is a push, the angle between the forces becomes 180° – 60° = 120°. Using the law of cosines:

$$R^2 = P^2 + Q^2 + 2PQ\cos(120°)$$
\begin{align} R^2 &= (20\text{ N})^2 + (15\text{ N})^2 \\ &+ 2 \times 20\text{ N} \times 15\text{ N} \times \cos(120°) \end{align}
\begin{align} R^2 &= 400\text{ N}^2 + 225\text{ N}^2 \\ &+ 600\text{ N}^2 \times (-0.5) \end{align}
$$R^2 = 400\text{ N}^2 + 225\text{ N}^2 – 300\text{ N}^2$$
$$R^2 = 325\text{ N}^2$$
$$R = \sqrt{325\text{ N}^2} = 18\text{ N}$$

Determining the Direction of the Resultant

Step 2: Using the tangent formula to find angle α (the angle the resultant makes with the 20 N force):

$$\tan\alpha = \frac{Q\sin(120°)}{P + Q\cos(120°)}$$
\begin{align} \tan\alpha &= \frac{15\text{ N} \times \sin(120°)}{20\text{ N} + 15\text{ N} \times \cos(120°)} \end{align}
\begin{align} \tan\alpha &= \frac{15\text{ N} \times 0.866}{20\text{ N} + 15\text{ N} \times (-0.5)} \end{align}
$$\tan\alpha = \frac{12.99\text{ N}}{20\text{ N} – 7.5\text{ N}}$$
$$\tan\alpha = \frac{12.99\text{ N}}{12.5\text{ N}} = 1.039$$
$$\alpha = \tan^{-1}(1.039) = 46.1° \approx 46°6’$$

Case (b) Results:

Magnitude of resultant force R = 18 N

Direction of resultant force = 46°6′ with the 20 N force

Explanation

Law of Cosines for Resultant Forces

When two forces P and Q act at a point with an angle θ between them, the magnitude of the resultant force R can be calculated using the law of cosines:

$$R^2 = P^2 + Q^2 + 2PQ\cos\theta$$

Direction of the Resultant

The direction of the resultant force is given by the angle it makes with one of the forces. This angle can be calculated using:

$$\tan\alpha = \frac{Q\sin\theta}{P + Q\cos\theta}$$

where α is the angle between the resultant and the force P.

Effect of Push vs. Pull

When both forces are pulls (or both are pushes), they act in the directions shown. However, when one force is a push and the other is a pull, the effective angle between them changes:

  • For case (a), both forces are pulls, so the angle between them is 60°.
  • For case (b), the 15 N force is a push, so it effectively acts in the opposite direction, making the angle between the forces 120° (180° – 60°).

Verification

The results can be verified by resolving the forces into horizontal and vertical components and then combining them:

Case (a):

Horizontal component: 20 N + 15 N × cos(60°) = 20 N + 7.5 N = 27.5 N
Vertical component: 15 N × sin(60°) = 12.99 N
Resultant magnitude: √(27.5² + 12.99²) = 30.4 N

Case (b):

Horizontal component: 20 N – 15 N × cos(60°) = 20 N – 7.5 N = 12.5 N
Vertical component: 15 N × sin(60°) = 12.99 N
Resultant magnitude: √(12.5² + 12.99²) = 18 N

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