Solve Trig Problem: 0.15kg Mass w/ Wall & Lift @ 65° Angle

[Tevion]

A mass 0.15kg sits in a corner between a vertical wall and a lift, that makes an angle of 65° with the wall. The only force acting on the mass is its weight (W), its normal reaction (Nr) from the lift and its normal reaction (Nw) from the wall. The mag of the acceleration due to gravity to be g = 10ms^2

1) Show that the angle between the directions of the forces W and Nr, is 155°

2) Draw a force diagram for the forces acting on the mass, giving the sizes of the angles between the forces

3) Draw a corrresponding triangle of forces, giving the sizes of angles

4) Use the triangle of forces to find the magnitudes of the two normal reactions in Newtons

Any help with the above would be most appriciated.

Thanks

 

[Mark44]

You're not likely to get much help without showing some attempt at these problems first.

 

[Tevion]

ok, its not great :( I am really struggling with this one.

Attached is my attempt

 

[Tevion]

Did that help?

 

[Tevion]

got it :D

 

[Mark44]

Good to hear. I was about to jump in with some help, but it looks like it's not needed now.

 

[Tevion]

Well i won't turn down assistance, and advice, so please go on :)

 

[Tevion]

now its a bit trickier and would like some help please.

rather than the angle being 65 degrees, but is x in radians where 0 < x < 1/2pi.

then Nr depend on the value of x.

the Nr from the lift have mag. fr(x) N. Assuming domain of F fr is (0,1/2pi]

1) we need to use the triangle of forces to show that f, fr has rule

[math]fr(x) = \frac{1.5}{sin x}[/math]

2) find the value of fr (1/2pi) and explain this value makes sense in context of model

 

[Mark44]

You've lost me on some of what you wrote.

the Nr from the lift have mag. fr(x) N. Assuming domain of F fr is (0,1/2pi]

You have too many symbols there and not enough words explaining what they mean. Some of it I understand from the previous problem.
N_r is the normal force of the mass. What does "mag. fr(x) N" mean? What does F fr mean?

Assuming that the only thing that has changed from the first problem is that the angle the lift makes with the wall is now x radians versus 65 degrees, the force diagram for the mass looks pretty much like the one for the first problem. The mass exerts a force straight down, its weight W. This vector can be decomposed into two normal forces: a horizontal force to the left, perpendicular to the wall, N_w, and a force that is perpendicular to the lift, N_r. The two normal forces N_w and N_r can be added to produce the force acting straight down, W.

Given that the lift makes an angle of x radians with the wall, the force diagram has one vector pointing to the left, and one vector pointing straight down, and one vector that points down and to the right. Vector W is the diagonal of a parallelogram.

Let's say that the angle between W and N_r is y radians. If the lift makes an angle of x rad. with the wall, it must be that x + pi/2 + y = pi, so y = pi - pi/2 - x = pi/2 - x radians.

If you have made a sketch of the situation, it should be fairly straightforward to calculate the magnitudes of the three vectors.