Question 20
A 150 kg crate is at rest on a ramp. The surface of the ramp is 3.2 m long and the top of the ramp is 1.1 m above the ground.
Calculate the normal force acting on the crate.
[2 marks] b)Determine the minimum coefficient of static friction required to keep the crate at rest.
[2 marks]The weight splits in two, and friction holds the part down the slope
The ramp and every force are drawn to scale. Watch the free-body diagram build up, then watch the ramp lifted from flat until friction runs out at exactly 20.11°.
The dashed arrows are the two components of the weight. They replace $mg$, they do not add to it. The mass and $g$ cancel in $F_f\div F_N$, so the coefficient depends only on the angle.
Reshape the ramp, then make the slips on purpose
Set the height of the ramp, the length of its surface and the mass of the crate. The ramp and the forces are drawn to scale. Then pick a slip to see the number it gives and why it is wrong. The question is a 1.1 m height, a 3.2 m surface and a 150 kg crate.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Part a) is the same in every method. Part b) has three routes, and each earns both marks.
- Weight $mg=150\times9.8=1470\text{ N}$, vertically downwards.
- Normal force $F_N=1380.38\text{ N}$, perpendicular to the ramp surface, away from it.
- Static friction $F_f=505.31\text{ N}$, parallel to the ramp surface, up the slope. It balances $F_\parallel=mg\sin\theta$.
The free-body diagram is shown for reference. It is not required for this question.
Find the angle of the ramp from the two lengths you are given.
$$\begin{aligned}\sin\theta&=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{1.1}{3.2}=0.34375\\\theta&=\sin^{-1}(0.34375)=20.11^\circ\end{aligned}$$
The ramp surface (3.2 m) is the hypotenuse and the height (1.1 m) is opposite the ramp angle.
The crate is at rest, so the forces perpendicular to the ramp balance.
$$F_N=F_\perp$$
The crate is stationary, so the forces perpendicular to the ramp are balanced. The normal force equals the component of the weight perpendicular to the ramp.
Marker: recognises that forces are balanced, therefore $F_\perp=F_N$
Resolve the weight perpendicular to the ramp, then substitute.
$$\begin{aligned}F_H&=F\cos\theta\\F_N=F_\perp&=mg\cos\theta\\&=150\times9.8\times\cos20.11^\circ\\&=1470\times\cos20.11^\circ\\&=1380.38\text{ N}\end{aligned}$$
Resolve the weight perpendicular to the ramp with $F_H=F\cos\theta$, taking the axis perpendicular to the ramp.
Allow FT for an incorrect angle.
Accept normal force range from 1380.38–1380.47 N inclusive. Also accept 1.38 kN.
Marker: correctly calculates the normal force acting on the crate
On the point of sliding, the forces parallel to the ramp balance and static friction is at its maximum.
$$\begin{aligned}F_f&=\mu_sF_N\\F_f&=F_\parallel\\\mu_sF_N&=mg\sin\theta\\\mu_s\,mg\cos\theta&=mg\sin\theta\\\mu_s&=\tan\theta\end{aligned}$$
The minimum $\mu_s$ is the one that leaves the crate just on the point of sliding, so the maximum static friction exactly balances the component of the weight down the ramp. With $F_f=\mu_sF_N$ this gives $\mu_s mg\cos\theta=mg\sin\theta$, which is the formula-book result for a ramp.
Accept $\mu_s=\tan\theta$.
Marker: correctly determines the method required to calculate the minimum coefficient of static friction
Substitute the ramp angle from Step 1.
$$\begin{aligned}\mu_s&=\tan\theta\\&=\tan20.11^\circ\\&=0.3661\\&=0.37\end{aligned}$$
Allow FT marks for errors in prior working.
Accept $\mu_s$ range from 0.36–0.37 inclusive.
Marker: correctly calculates the minimum coefficient of static friction
As in Method 1. The angle is $\theta=20.11^\circ$, the forces perpendicular to the ramp balance, and the normal force is the part of the weight into the ramp.
$$\begin{aligned}F_N=F_\perp&=mg\cos\theta\\&=150\times9.8\times\cos20.11^\circ\\&=1380.38\text{ N}\end{aligned}$$
Marker: recognises that forces are balanced, therefore $F_\perp=F_N$
Marker: correctly calculates the normal force acting on the crate
On the point of sliding, static friction balances the component of the weight down the ramp. Resolve the weight parallel to the ramp.
$$\begin{aligned}F_V&=F\sin\theta\\F_\parallel&=mg\sin\theta\\&=150\times9.8\times\frac{1.1}{3.2}\\&=505.31\text{ N}\\F_f&=F_\parallel=505.31\text{ N}\end{aligned}$$
You already know $\sin\theta=\dfrac{1.1}{3.2}$ exactly from Step 1, so you can use the ratio and skip the rounded angle.
Marker: correctly determines the method required to calculate the minimum coefficient of static friction
Rearrange the formula-book friction relation and divide.
$$\begin{aligned}F_f&=\mu_sF_N\\\mu_s&=\frac{F_f}{F_N}=\frac{505.31}{1380.38}\\&=0.37\end{aligned}$$
Method 2: $F_\parallel=mg\sin\theta=150\times9.8\times\dfrac{1.1}{3.2}=505.31$ N, $\mu_s=\dfrac{F_f}{F_N}=\dfrac{505.31}{1380.38}=0.37$.
Allow FT marks for errors in prior working.
Accept $\mu_s$ range from 0.36–0.37 inclusive.
Marker: correctly calculates the minimum coefficient of static friction
As in Method 1. The angle is $\theta=20.11^\circ$, the forces perpendicular to the ramp balance, and the normal force is the part of the weight into the ramp.
$$\begin{aligned}F_N=F_\perp&=mg\cos\theta\\&=150\times9.8\times\cos20.11^\circ\\&=1380.38\text{ N}\end{aligned}$$
Marker: recognises that forces are balanced, therefore $F_\perp=F_N$
Marker: correctly calculates the normal force acting on the crate
On the point of sliding, $\mu_s=\tan\theta$ (Step 4 of Method 1). Tangent needs the side adjacent to the angle, the horizontal run, so find it with Pythagoras.
$$\begin{aligned}c^2&=a^2+b^2\\\text{run}&=\sqrt{3.2^2-1.1^2}\\&=3.00\text{ m}\end{aligned}$$
The 3.2 m surface is the hypotenuse, not the run. The run is the third side of the right-angled triangle, under the ramp.
Marker: correctly determines the method required to calculate the minimum coefficient of static friction
Divide the height by the run.
$$\begin{aligned}\mu_s=\tan\theta&=\frac{\text{opposite}}{\text{adjacent}}\\&=\frac{1.1}{3.00}\\&=0.37\end{aligned}$$
Method 3: horizontal run $=\sqrt{3.2^2-1.1^2}=3.00$ m, $\mu_s=\tan\theta=\dfrac{1.1}{3.00}=0.37$.
Allow FT marks for errors in prior working.
Accept $\mu_s$ range from 0.36–0.37 inclusive.
Marker: correctly calculates the minimum coefficient of static friction
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Treating 3.2 m as the horizontal run. You get $\theta=\tan^{-1}\!\left(\dfrac{1.1}{3.2}\right)=18.97^\circ$, which gives $F_N=1390.16\text{ N}$ and $\mu_s=0.34$. The 3.2 m is the surface, the hypotenuse.
- Using $\sin\theta$ instead of $\cos\theta$ for the normal force. You get $1470\times0.34375=505.31\text{ N}$. That is the component of the weight down the ramp, not into it.
- Your calculator in radian mode. You get $1470\times\cos(20.11)=448.93\text{ N}$. Check the mode before you start.
- Using $g=9.81$. You get $F_N=1381.79\text{ N}$, outside the accepted range of 1380.38 to 1380.47 N. Use $g=9.8$ from the formula book.
- Inverting the ratio in b). You get $\dfrac{F_N}{F_\parallel}=\dfrac{1380.38}{505.31}=2.73$. The coefficient is friction divided by the normal force, so on this gentle ramp it must be less than 1.
The past QCAA questions this one was modelled on
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it