A wheelbarrow inclined at 60º to the horizontal is pushed with a force of 150N. What is the horizontal component of the applied force
- 130N
-
75N
- 60N
- 50N
The correct answer is: B
Explanation
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 .