A wheelbarrow inclined at 60º to the horizontal is pushed with a force of 150N. What is the horizontal component of the applied force
The horizontal component F\(_x\) can be calculated using the formula: F\(_x\) = F⋅ cos (\(\theta\))
F = 150N and \(\theta\) = 60º
F\(_x\) = 150 x 0.5 = 75N (cos 60º = 0.5)
There is an explanation video available below.
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