Question 22

Draw the parcel as a block on a ramp at $25^\circ$ to the horizontal, with three forces drawn from its centre.

• $F_g=117.6\text{ N}$, vertically downwards.
• $F_N=106.58\text{ N}$, perpendicular to the ramp surface, pointing away from it.
• $F_f=31.97\text{ N}$, parallel to the ramp, pointing up the slope. It opposes the sliding.

The components of $F_g$ may be shown dashed, because they are not extra forces: $F_\parallel=49.70\text{ N}$ down the slope and $F_\perp=106.58\text{ N}$ into the ramp. The angle between $F_g$ and the line perpendicular to the ramp is $25^\circ$.

Start from $F=ma$ with $a=g$.

$$\begin{aligned}F&=ma\\F_g&=mg\\&=12\times9.8\\&=117.6\text{ N}\end{aligned}$$

The parcel does not move perpendicular to the ramp, so the forces across the ramp balance and $F_N=F_\perp$. Resolve $F_g$ along the perpendicular with $F_H=F\cos\theta$.

$$\begin{aligned}F_H&=F\cos\theta\\F_N=F_\perp&=F_g\cos\theta\\&=117.6\times\cos25^\circ\\&=106.58\text{ N}\end{aligned}$$

The parcel is sliding, so use the coefficient of kinetic friction, $\mu_k=0.30$.

$$\begin{aligned}F_f&=\mu F_N\\&=0.30\times106.58\\&=31.97\text{ N}\end{aligned}$$

Resolve $F_g$ along the ramp with $F_V=F\sin\theta$.

$$\begin{aligned}F_V&=F\sin\theta\\F_\parallel&=F_g\sin\theta\\&=117.6\times\sin25^\circ\\&=49.70\text{ N}\end{aligned}$$

Take down the ramp as positive. Friction acts up the ramp, so it goes in with a minus sign.

$$\begin{aligned}F_{net}&=F_\parallel-F_f\\&=49.70-31.97\\&=17.73\text{ N}\end{aligned}$$

Divide the net force by the mass of the parcel.

$$\begin{aligned}F&=ma\\a&=\frac{F_{net}}{m}\\&=\frac{17.73}{12}\\&=1.4775\text{ m s}^{-2}\end{aligned}$$

The free-body diagram and the forces are the same as in Method 1. Draw $F_g$, $F_N$ and $F_f$ with friction up the ramp, then resolve and add along the ramp.

$$\begin{aligned}F_g&=mg=12\times9.8=117.6\text{ N}\\F_N=F_\perp&=F_g\cos\theta=117.6\times\cos25^\circ=106.58\text{ N}\\F_f&=\mu F_N=0.30\times106.58=31.97\text{ N}\\F_\parallel&=F_g\sin\theta=117.6\times\sin25^\circ=49.70\text{ N}\\F_{net}&=F_\parallel-F_f=49.70-31.97=17.73\text{ N}\end{aligned}$$

Skip the acceleration. Name the principle instead: the work done by the net force over the length of the ramp equals the change in kinetic energy of the parcel. The parcel starts from rest, so $u=0$ and the starting kinetic energy is zero. Use $F_{net}=17.73\text{ N}$, $s=4.5\text{ m}$ and $m=12\text{ kg}$.

$$\begin{aligned}W&=Fs\qquad KE=\tfrac12mv^2\\F_{net}\,s&=\tfrac12mv^2-\tfrac12mu^2\\17.73\times4.5&=\tfrac12\times12\times v^2-0\\79.785&=6v^2\\v^2&=13.2975\end{aligned}$$