The attached shows the angle to be 65° , but that now is $\displaystyle x$ in radians where $\displaystyle 0 < x < \frac{1}{2}\pi$.

Then the normal reaction depend on $\displaystyle x$ ; we let the normal reaction from the lift have magnitude $\displaystyle fr(x)$ newtons. We assume that the domain of $\displaystyle f$ $\displaystyle fr$ is $\displaystyle (0,\frac{1}{2} \pi]$

1) Use a triangle of force to show that the $\displaystyle f$ has the rule

$\displaystyle fr(x) = \frac{1.5}{sin x}$

2) find the value of $\displaystyle fr(\frac{1}{2} \pi)$ and explain why this value makes sense in situation modelled

2. It's not clear from the figure, but I'm going to guess that the ball m has a weight of 1.5N. The force of gravity is pulling down on the ball, and its weight is counter-acted by the normal force of the ramp. That normal force acts at angle x. There is also a third force at play here - the reaction force from the side wall, which I will call $\displaystyle F_s$. See the attached figure - it shows how all three forces fit together to balance. From this you can see that $\displaystyle W = F_R sin(x)$, so $\displaystyle F_R = 1.5/sin(x)$ .

2. If $\displaystyle x = \pi/2$, you have the situaton where the ball is simply resting on a table, so of course the normal force is equal to the weight.

3. ok great - thank you for that. So how can we draw this as a graph. i.e $\displaystyle y = fr(x)$ with a table of values?

Then from that graph, explain why $\displaystyle fr$ has an inverse function? leading from there - whats its domain and image set? can we then draw a graph?

4. You already know that the function of the reaction force for a given angle x is is f = 1.5N/sin(x). Make up a table of x and f where x varies from 0 to 90 degrees and see how f behaves. You can then plot f versus x.

To find the inverse function of f=1.5/sin(x) you need to rearrange this to get x by itself:

f = 1.5N/sin(x)
sin(x) = 1.5N/f
x = arcsin(1.5N/f)

This tells you what the angle x is to achieve a given reaction force f. The domain of 1.5N/f is limited to values less than or equal to 1 (since the arcsin of a value greater than 1 is not allowed). So if 1.5N/f must be less than or equal to 1, that means that f must be geater than or equal to 1.5N.

5. Any help on what the domain and image set is?

not $\displaystyle (0, \frac{1}{2} \pi]$