Answer:
[tex]P = 415.037\,N[/tex]
Explanation:
Let assume that direction of pull force is parallel to the ramp. The equations of equilibrium for the crate are: (x' is the axis parallel to the ramp, y' is axis perpendicular to the ramp)
[tex]\Sigma F_{x'} = P - m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0[/tex]
The force exerted by the man on the crate can be calculated after manipulating the equation of equilibrium algebraically:
[tex]P = m \cdot g \cdot \sin \theta + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]
[tex]P = m\cdot g \cdot (\sin \theta + \mu_{k}\cdot \cos \theta)[/tex]
[tex]P = (50\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\sin 30^{\textdegree} + 0.40 \cdot \cos 30^{\textdegree})[/tex]
[tex]P = 415.037\,N[/tex]
