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?
Hello, classicstrings!
I drew a sketch for #2 . . . and saw an "eyeball" solution!
Code:* * * * * b * * a+b * * * 120* * * * * * * * * * * * a
We are given: The angle between and is and
We have an isosceles triangle with equal sides and vertex angle
Therefore, the base angles are
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:
where m = 50 kg.
Thus a = 4.9 m/s^2.
-Dan
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:
<--Make sure you understand why this is cosine, not sine!
There is no acceleration in the y direction so .
or N = 130 N (keeping 2 significant figures.)
-Dan