Some Random Questions....

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• October 28th 2006, 07:58 AM
classicstrings
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?
• October 28th 2006, 08:44 AM
CaptainBlack
Quote:

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
• October 28th 2006, 09:19 AM
Soroban
Hello, classicstrings!

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

Quote:

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

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 $30^o.$

• October 28th 2006, 11:16 AM
topsquark
Quote:

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
• October 28th 2006, 11:21 AM
topsquark
Quote:

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