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Resolve a weight of 10N in two direction...

Resolve a weight of 10N in two directions which are parallel and perpendiular to a slope inclined at `30^@` to the horizontal.

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To resolve a weight of 10 N into two components that are parallel and perpendicular to a slope inclined at \(30^\circ\) to the horizontal, we can follow these steps: ### Step 1: Understand the Problem We have a weight \(W = 10 \, \text{N}\) acting vertically downwards. We need to resolve this weight into two components: one that is parallel to the slope and one that is perpendicular to the slope. ### Step 2: Identify the Angles The slope is inclined at \(30^\circ\) to the horizontal. Therefore, the angle between the weight and the line perpendicular to the slope is \(90^\circ - 30^\circ = 60^\circ\). ...
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