Angular velocity and rotational equilibrium

[mollybethe]

Homework Statement

A solid rectangle of uniform density has one corner at the origin. It has a mass of 50 kg. The rectangle
has a length of 4 m in the z-direction, a length of 3 m in the y-direction, and a length of 2 m in
the x-direction. The pivot is at the center of mass.

There is a 50 N force in the x = y = z direction acting on the corner located at the origin. There is
a 35 N force acting at the coordinates, (0.5, 1, 0), which is in the z-direction. There is also a 60 N
force acting at the coordinates, (2, 2, 1), which makes an angle of 130 degrees to the x-axis and
is at an angle of 115 degrees to the z-axis.
(a.) What is the angle between the 50 N and the 60 N forces?
(b.) What will be the angular acceleration of this object?
(c.) If you were to apply a force on the surface of the prism solely in the z-direction in order
to keep it in rotational equilibrium, pick a place to apply the force and find the
magnitude of it.

Homework Equations

The Attempt at a Solution

I know this is a Torque problem and I need to sum the forces on part c, but I have no idea how to determine the angle that is being indicated. Have never had anything in 3 dimensions before and I can't find a single problem like it in my textbook to understand how to do it.

 

[tiny-tim]

hi mollybethe!

(c.) If you were to apply a force on the surface of the prism solely in the z-direction in order to keep it in rotational equilibrium, pick a place to apply the force and find the magnitude of it.
…
… I need to sum the forces on part c, but I have no idea how to determine the angle that is being indicated.

the direction and magnitude of the sum is found the usual way

the line of application of the sum is found by using moments (about any point)

(and to find an angle, use the dot product )

 

[mollybethe]

This is pretty much a shot in the dark. I have never done anything in 3-dimensions We only covered this in class for a bit. The instructor tried to jam 3 chapters into one class, so here we go
a.
cos$$\theta$$=0
$$\theta$$=90

b.∑ τ =I$$\alpha$$
$$\tau$$=r x F sin $$\theta$$

I=(1/12)(50)(2²+3²)=54.17

∑ τ=(4)(50)sin 0+(2)(35)sin 63+(1.5)(60)sin130=193.37

$$\theta$$63 was found by taking tan^-1(1/.5), can I do that?

193.37=54.17$$\alpha$$
$$\alpha$$=3.57m/s²

c.I am so lost, I can't do three dimensions.

 

[tiny-tim]

hmm … let's try (a.) first …

There is a 50 N force in the x = y = z direction acting on the corner located at the origin.
…
There is also a 60 N force acting at the coordinates, (2, 2, 1), which makes an angle of 130 degrees to the x-axis and is at an angle of 115 degrees to the z-axis.
…
(a.) What is the angle between the 50 N and the 60 N forces?

call the unit vectors in the directions of the two forces p and q

you know that p is parallel to 2i + 2j + k

and that q.i = cos130° and q.k = cos115°

(so q.j = √(1 - cos²130° - cos²115°)) …

now what is p.q ? ​

 

[mollybethe]

I am really trying to understand what you are doing, but I haven't done three dimensional analysis before, I am only through Calculus 1...that was the only prereq. for this course, so bare with me...I assumed you named them respectively, so why is p parallel to 2i + 2j + k?

I sent my professor an message and he said to just use trig, but I don't see it. The only formula I have is A $$\bullet$$B=ABcos$$\theta$$. If it is at the origin wouldn't the coordinate be (0,0,0) so wouldn't that give me a 0 dot product?

 

[tiny-tim]

hi mollybethe!

I am really trying to understand what you are doing, but I haven't done three dimensional analysis before, I am only through Calculus 1...that was the only prereq. for this course, so bare with me...I assumed you named them respectively, so why is p parallel to 2i + 2j + k?

i'm sorry, my eye must have skipped a line

p is parallel to i + j + k

I sent my professor an message and he said to just use trig, but I don't see it. The only formula I have is A $$\bullet$$B=ABcos$$\theta$$.

(LaTeX for dot is \cdot )

ah, you should also learn the formula A.B = A_xB_x + A_yB_y + A_zB_z

(and of course i.i = j.j = k.k = 1,
i.j = j.k = k.i = 0)