Using the diagram above, the effective force pushing it forward at an angle 60º is
The effective force pushing it forward is the horizontal component of the force = F\(_x\) = F Cos\(\theta\)
F\(_x\) = F Cos\(\theta\) = 50 x Cos 60º = 50 x 0.5 = 25N
There is an explanation video available below.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}