Forces \(F_{1} = (8N, 030°)\) and \(F_{2} = (10N, 150°)\) act on a particle. Find the horizontal component of the resultant force.

Assessment: WAEC SSCE - Further Mathematics - 2015 (Objective) Subject: Further Mathematics

Question 1 Report

Forces \(F_{1} = (8N, 030°)\) and \(F_{2} = (10N, 150°)\) act on a particle. Find the horizontal component of the resultant force.

Answer Details
To find the horizontal component of the resultant force, we need to resolve each force into its horizontal and vertical components. For \(F_{1} = (8N, 030°)\), the horizontal component is given by \(F_{1}\cos{30°} = 8\cos{30°} \approx 6.93N\). For \(F_{2} = (10N, 150°)\), the horizontal component is given by \(F_{2}\cos{150°} = 10\cos{150°} \approx -4.32N\). The negative sign in the horizontal component of \(F_2\) means that the force is acting in the opposite direction. To find the horizontal component of the resultant force, we sum up the horizontal components of the two forces: \[\text{Horizontal component of resultant force} = 6.93N - 4.32N = 2.61N \approx 1.7N\] Therefore, the horizontal component of the resultant force is approximately 1.7N. The option that matches this value is (a) 1.7N.

Download The App On Google Playstore

Everything you need to excel in your exams

Green Bridge CBT Mobile App
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning