Answer:
Frictional force (F(frict)) = 17.1 N
Normal force (F(norm)) = 78.0 N
Gravitational force (F(grav)) = 63.7 N
Acceleration (a) = 1.2 ms⁻²
Step-by-step explanation:
Newton's second law:
[tex]\boxed{F_{net}=ma}[/tex]
where:
Use trigonometry to resolve the 28.5 N force into its horizontal and vertical components since it is acting on the particle at 30°:
[tex]\implies F_x=28.5 \cos 30^{\circ}[/tex]
[tex]\implies F_y=28.5 \sin 30^{\circ}[/tex]
Due to the particle's mass, m, and the acceleration due to gravity, g:
[tex]\boxed{W=mg}[/tex]
Given:
[tex]\begin{aligned}W & = mg\\\implies W & = 6.5 \times 9.8\\W & = 63.7\;\sf N\end{aligned}[/tex]
The particle is moving parallel to the plane (the horizontal), so acceleration perpendicular to the plane is zero.
Resolving forces vertically (taking ↑ as positive):
[tex]\begin{aligned}F & = ma\\\implies R-W-28.5 \sin 30^{\circ}&=6.5 \times 0\\R-W-28.5 \sin 30^{\circ}&=0\\R & = W+28.5 \sin 30^{\circ}\\R & = 6.5(9.8)+28.5(0.5)\\R & = 63.7+14.25\\R & = 77.95\\R & = 78.0 \;\; \sf N\; (1\:d.p.)\end{aligned}[/tex]
When a moving object is acted on by a frictional force, fiction is limiting, and the frictional force F is at its maximum value:
[tex]\boxed{F = \mu R}[/tex]
(Where μ is the coefficient of friction and R is the normal reaction).
[tex]\begin{aligned}F & = \mu R\\\implies F & = 0.22 \times 77.95\\F & = 17.149\\F & = 17.1 \;\; \sf N\;(2\:d.p.)\end{aligned}[/tex]
Friction always acts in the opposite direction to motion (or potential motion).
Resolving forces horizontally (taking → as positive):
[tex]\begin{aligned}F & = ma\\\implies 28.5 \cos 30^{\circ}-F & = 6.5a\\28.5 \left(\dfrac{\sqrt{3}}{2}\right)-17.149 & = 6.5a\\\dfrac{57\sqrt{3}}{4}-17.149 & = 6.5a\\7.532724008 & = 6.5a\\a & = 1.158880617\\a & = 1.2\; \sf ms^{-2}\;\;(1\:d.p.)\end{aligned}[/tex]
