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Question 17
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(a) A particle Q is acted upon by three coplanar forces 40N, 15√3N, and 20N, as
shown in the diagram
Find the force required to prevent Q from moving.
(b) A body of mass 5 kg at rest is acted upon by forces F1 = (10N,0900),
F2 = (20N, 210°) and F3 = (4N, 330°). Find, correct to one decimal place, the
magnitude and direction of the resultant force.
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This question was on resolution of coplanar forces and candidates' performance was
reportedly poor. However, candidates performed better in part (a) than in part (b).
Teachers were encouraged not to relent in their effort to teach this topic and WAEC was
encouraged to set questions repeatedly from this section of the syllabus so as to
encourage both teachers and candidates to prepare adequately for the examination.
In part (a), candidates were expected to resolve the forces both horizontally and
vertically. Resolving horizontally, we have 40Cos 30° + 15√3 = 20√3 + 15√3 =
35√3. Resolving vertically, we have 40Sin30° - 20 = O. Therefore,magnitude of resultant force =√(35√3)² +0² = 35√3N. Since the force had its vertical component = 0,
it implies that the required force will act in a direction opposite the horizontal component
= 270°, i.e required force =( 35√3, 270°) .
In part (b), Forces F1 , F2 and F3 could be represented diagrammatically as
Horizontal component of Resultant force = 10 - 20Cos60° - 4Cos60° = -2.
Vertical component of Resultant force = 4Sin60° - 20Sin60° = -13.8564.
Resultant force, R =√(_2)² + (-13.8564)² = 14.0N. The direction of the resultant
force = 270° - tan–¹ [-13.8564) ²= 188.2°.
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