So for the deflection I found a formula P.L^3/(48.E.I)
The right answer should be 3,33cm
I calculated Ix and put it there to the formula and what I got is 3,39cm.
Is my work at least a bit correct?
I decided to solve it on my own. But I have one question. How could I calculate the Moment of Area of the cross-section of the beam? That's the only thing I have no idea what it is and how to calculate it.
I'm a high school student and we made this experiment at school and I just wanted to know if I could calculate it with some formula. But I think that I don't have enough knowledge to solve it by myself. Is there anybody who would like to solve it for me please?
The beam is supported on the ends of it.
Length of it is 100cm. The other two dimensions are 5cm and 0,5cm. The cross section is rectangular.
I'm a bit confused. Hopefully I wil somehow solve it.
I tried to use F/S = E.ε to find the new length after deformation and than use pytagoras theorem in right angled triangle to find how much did the middle point lower.
But it was said to me that my method is wrong and that there exists some formula to solve it. Can you please tell me where could...
Homework Statement
Good evening,
I have a homework where the question is "How high will person jump on Mars?"
Homework Equations
v2 = 2.amars.h1
V2 = 2.g.h2
The Attempt at a Solution
I know everything to solve it. I can solve it by h1 / h2 . In this created equation the only unknown is...
Hello, i have a sequence {1,2,13,62,313...} and I have to find out the rule for n-th number. I've found out that every next number is five times bigger but then is added or subtracted 3. For example 1x5 -3 = 2 and 2x5 +3 = 13 and so on. Can you please give me some advice how to create the...
Wait a second please. I'm confused.
My current opinion is that a shooted bullet has to have greater acceleration because it will fall down faster than an object which was released from rest.
Bullet is in my opinion pushed by the gun thus he has to have greater acceleration.
Please can you...
Why? Because I think that it's acceleration has to be greater than g because the second stone will "catch" the first one.
There is nothing written there about initial velocity.
Homework Statement
A man let a stone from height 49 metres. After 1 second he threw a second stone. Both stones fell on same time. What is the velocity of second stone?
Homework Equations
h=1/2gt^2
h=1/2a(t-1)^2
The Attempt at a Solution
So I assume that if he let the first stone then the...
Hello,
I have a little problem. I try to find out why the second equation in the picture is correct. How can I get it? Down on the paper is what I did.
Where did the (1+V2/V1) disappear from the right side of equation?
Thank You for every help.
So basically if I understood everything correctly.
Tension of the rope is compensating weight in every position?
Does it mean that at 6 o'clock position if we had given the max tension of the rope we can calculate in what speed will the rope tear up right?
The equation would be like this...
So am I right if I say that there are only weight and force of the rope?
How is then possible that the object won't stop "in the 6am position"? Because it's too fast?
Homework Statement
Imagine a vertically placed clock. What forces have an effect on the object when it is in the position of 3am?
Homework Equations
No need od equations i guess.
The Attempt at a Solution
So I figured out that when the object is in the position of 12am then weight of the...
Nope. I can't figure anything out. I will make a break for a while and then I will try it again because now I can't do it.
Thank You you for everything. I really appreciate it.
What I did this time is this.
vn = vn-1 .atn
which means a=(vn -vn-1) / tn
and that's what I put in equation s=vn-1.tn + 1/2.a.tn2
what I got is this tn = 2s/(vn + vn-1)
Can this somehow help me? I'm really sorry for my stupidness but I can't figure it still out.