Answer to Question #165895 in Classical Mechanics for Safaa gulzar

Question

a 10kg box sits on a slope at an angle of 15 degrees:

What is the normal reaction force the slope exerts on the box.

What is the component of the weight parallel to the slope.

If The box accelerates down the slope at 0.5m/s2, calculate the magnitude of friction.

Answer 1

$\text{the following forces act on the box:}$

$\vec{F_g} = mg = 10*9.8=98 \text{ gravity force}$

$\vec{N}-\text{reaction force }$

$\vec{F_{fr}}\text{ friction force}$

$\text{Let us choose the coordinate system - the X axis parallel to the slope}\newline \text{ and the Y axis perpendicular to the slope}$

$\text{projection of forces on the Y axis:}$

$\vec{N}-\vec{F_{gy}}=0$

$\vec{F_{gy}}=\vec{F_g}\cos\alpha= mg\cos{15\degree}\approx94.66\ N$

$\vec{N}=\vec{F_{gy}}=94.66\ N$

$\text{projection of forces on the X axis:}$

$\vec{F}= \vec{F_{gx}}-\vec{F_{fr}}$

$\vec{F}=ma=10*0.5=5\ N$

$\vec{F_{gx}}= mg\sin\alpha=10*9.8*\sin15\degree\approx25.36\ N$

$\vec{F_{fr}}= \vec{F_{gx}}-\vec{F}=25.36-5=20.36\ N$

Answer: 94.66 N normal reaction force;25.36 N component of the weight parallel to the slope;

20.36 N magnitude of friction