1. ## Some Random Questions....

Hello! I'm having some difficulty with these questions. Thanks for the help!

1.

If a = 2j - 3k
b = 3i - 3j + ck

and

abs(a+b) = root35

Then c is equal?

2.

The angle between vectors a and b is 120 and abs of a = abs of b. The angle b/w a and a + b is?

3.

A woman of mass 50kg stands in a lift which is moving upwards with acceleration of am/s^2. THe floor of the lift exerts a force of magnitude 75g Newtons on the women. Then value of a is?

4.

A body of mass 15kg slides down a straight slide which is inclined at an angle of 30 degrees to horizontal. The normal force, in Newtons of the slide acting on the boy is?

2. Originally Posted by classicstrings
1.

If a = 2j - 3k
b = 3i - 3j + ck

and

abs(a+b) = root35

Then c is equal?
a+b=3i + (2-3)j + (-3+c)k = 3i -j+(c-3)k

so

|a+b|^2 = 3^2 + 1 + (c-3)^2 = c^2 - 6c + 19

so if |a+b|=sqrt(35) then:

c^2 - 6c + 19 = 35,

or:

c^2 - 6c -16 = 0

which factorises by inspection to:

(c-8)(c+2)=0,

so c=8 or -2.

RonL

3. Hello, classicstrings!

I drew a sketch for #2 . . . and saw an "eyeball" solution!

$\text{2. The angle between vectors }\vec{a}\text{ and }\vec{b}\text{ is }120^o,\;\text{ and }|a| = |b|.$
$\text{Find the angle between }\vec{a}\text{ and }\overrightarrow{a + b}$
Code:
      *
*  *
*     *
b *        * a+b
*           *
* 120*         *
* * * * * * * * * *
a

We are given: The angle between $\vec{a}$ and $\vec{b}$ is $120^o$ and $|a| = |b|.$

We have an isosceles triangle with equal sides $|a|$ and vertex angle $120^o.$

Therefore, the base angles are $30^o.$

4. Originally Posted by classicstrings
3. A woman of mass 50kg stands in a lift which is moving upwards with acceleration of am/s^2. THe floor of the lift exerts a force of magnitude 75g Newtons on the women. Then value of a is?
When in doubt, use Newton's 2nd. I have a Free-Body Diagram for the woman. There is a normal force, N, acting upward, she has a weight, w = mg, acting downward. The lift the woman is standing in is accelerating upward at a m/s^2, so she is as well. I have a +y direction in the upward direction.

The floor of the lift is exerting a force of 75g N on her. This is the normal force, so N = 75g N. Thus Newton's 2nd on the woman reads:
$\sum F_y = N - w = ma$

$75g - mg = ma$

$a = \frac{1}{m} (75g - mg)$ where m = 50 kg.

Thus a = 4.9 m/s^2.

-Dan

5. Originally Posted by classicstrings
4. A body of mass 15kg slides down a straight slide which is inclined at an angle of 30 degrees to horizontal. The normal force, in Newtons of the slide acting on the boy is?
I have a Free-Body Diagram of the boy. There is a normal force, N, acting perpendicularlly out of the slide, there is a weight, w = mg, acting straight down. Friction is not mentioned in the problem so I will ignore it. I am going to set a +x direction down the slide and a +y direction in the direction of the normal force.

Newton's 2nd in the +y direction says:
$\sum F_y = N - w \cdot cos(30) = ma_y$ <--Make sure you understand why this is cosine, not sine!

There is no acceleration in the y direction so $a_y = 0$.

$N - mg \cdot cos(30) = 0$

$N = mg \cdot cos(30) = 127.306 \, N$

or N = 130 N (keeping 2 significant figures.)

-Dan