Question

Availing medals ... Please help.............

Answer 1

I thought it was spanish at first but I can't understand it so it must be portugues.

Answer 2

Por favor, se concentrar na "Carga de tijoles" e como melhor você pode desenhar um diagrama de corpo livre para ele. Você já desenhado diagramas de corpo livre antes?

Answer 3

A brick load of 15.0 kg is suspended by the end of a rope that passes over a small pulley without friction. A balance of 28.0 kg is stuck on the other end of the rope, as shown in Figure 5:51. The system is released from rest.

a) Draw a free body diagram for the load of bricks and the other for balance.
b) What is the magnitude of the upward acceleration of the load of bricks?
c) What is the tension in the rope during movement of the load? How this tension is related to the load? Since this voltage is related to the balance?

Answer 4

Certamente poderíamos voltar a esta questão mais tarde!

Answer 5

ok...

Answer 6

ara aplicar as equações de equilíbrio, é necessário fazer um Diagrama de Corpo Livre (DCL) do objeto a ser estudado. O DCL é um esquema onde fazemos a análise do corpo isoladamente, isto é, livre de vínculos físicos. Para isso, devem ser representadas todas as forças e momentos que atuam sobre o corpo a ser estudado.

Answer 7

@hoblos

Answer 8

for the first question you have to draw each object alone and show the forces applied on it

do you know what are the forces applied on each ?

Answer 9

acceleration and low acceleration above?

Answer 10

they are the weight of the object, and the tension in the rope

Answer 11

For this kind of problems, you have to be able to:
- identify the forces acting on each body
- relate the velocities of the different parts to one another
- apply laws of motion

Answer 12

what do I do now?

Answer 13

Have you done the 3 things I wrote a fow minutes ago?
Show us your diagram.

Answer 14

each of the 2 objects is having a weight force ( related to its mass) and a tension force in the rope

Answer 15

what is the weight of each ?

Answer 16

28kg e 15kg os pesos....

Answer 17

?

Answer 18

Weight and mass are different. It is the weight that is has mechanical action.

Answer 19

Now you have to apply Newton's laws of motion.

Answer 20

Aceleraçao de baixo de 28kg:
F=ma
28g-T=28a

Answer 21

?

Answer 22

Correct, now do the same thing for the other mass.

Answer 23

Aceleraçao de baixo de 28kg:
F=ma
28g-T=28a

Answer 24

o que faço agora?

Answer 25

You have already written this. I was talking about the 15 kg mass.

Answer 26

Aceleraçao de cima de 15kg:
F=ma
T-15g=15a

Answer 27

e agora?

Answer 28

Perfect, solve the system for a and you have the answer you are looking for.

Answer 29

but now what do I do?

Answer 30

?!?!? Can you understand my post?

Answer 31

yes, but now I do not know what to do anymore, this is already the answer? or need to do anything else?

Answer 32

I wrote: "solve the system for a and you have the answer you are looking for."

Just do that.

Answer 33

If you do this, this will give you the answer to question b).

Then you will solve for T, which will give you the answer to question c).

Answer 34

tried here, but could not: (

Answer 35

have to share?