Several problems put together in one thread, dynos, density etc.

[Fishingaxe]

Homework Statement

Problem A)

The brick A has a mass of 2.0kg and the brick B has a mass of 3.0kg. As you can see they are bound together via a dyno and is draged by the constant speed v = 2.5m/s
The dynos force never changes, meaning they will always have the force 12 N and 7 N.

How big is the frictionforces that acts upon the bricks under the movement of those same bricks?



Problem B) The dynos each weigh 0.15kg and the weights each weigh 0.35kg



What does the dynos show?

Problem C) An empty beaker weighs 67.0g, when 100 ml fluid is poured into that beaker it suddenly weighs 153.3g. Calculate the fluid's density.

Problem D) 200 sheets of paper weighs 1.0kg, the length and width of the papers are 29.6 cm and 21.0 cm and the height of those papers is 16 mm. What density does these papers have?

Problem E) A container with a circular bottom is partly filled with ethanol(see picture)
Calculate the liquid pressure at the container's bottom.


Problem F) The water pressure in a waterpipeline(not sure if this is the correct translation) at a house is 200 kPa. The mouth of the water tap has an area of 0.5cm².
With how much power do you have to press your finger against the opening if you open the water tap fully and try to resist the water power. (I hope this one makes sense to some of you, my teacher had left out some words so I had to fill into the best of my ability).

Last problem G) The figure shows a cuboid of iron. You want to set the iron block on a surface so that the pressure against the surface becomes as large as possible. Determine the maximum pressure.

The Attempt at a Solution

Problem C solution: P = m/V. 153.3 - 67 = 86.3 g. 100 ml of fluid weighs 86.3 g, P = m/V = 86.3/100 = 0.863 g/ml. is this the correct density?

Problem D solution. P = m/V. m = 1kg(1000g) V = width * length * height = 29.6 * 21 * 1.6 = 994.56cm^3.

1000/994.56 = 1.005. The density is 1.005g/cm^3. Is this also correct?

I am very sorry to say that I am clueless with the other problems, any comments that will put me to the right track of solving these is extremely appreciated.

I just want to take a few sentences and say that this seems to be a great forum with a lot of intelligent people which is rare to find, I am extremely grateful that a forum such as this is open to everybody. This forum even has a special place where you can get help with your homework no matter how simple/hard. Those people who helps people here actually plays a larger part than the teachers in a lot of ways and I thank you all.

 

[mfb]

C/D look right.

Hints:
A/B: Which forces act on A/2? Does the velocity of A/2 change? What can you conclude based on that?
Afterwards, do the same for B/1.

E: I am sure you have some formula how to calculate such a pressure in a liquid.

With how much power do you have to press your finger against the opening if you open the water tap fully and try to resist the water power.

Power does not make sense, I guess force is meant.

Spoiler

How are pressure and force related?

Note: that is a question you should find yourself.

G should be easy if you find the solution for F.

 

[Fishingaxe]

C/D look right.

Hints:
A/B: Which forces act on A/2? Does the velocity of A/2 change? What can you conclude based on that?
Afterwards, do the same for B/1.

E: I am sure you have some formula how to calculate such a pressure in a liquid.

Power does not make sense, I guess force is meant.

Spoiler

How are pressure and force related?

Note: that is a question you should find yourself.

G should be easy if you find the solution for F.

Thank you for your comment, I did solve the problem B I think.

Solution: The total mass of the dynos + weights is 1kg, the dyno 1 carries the weight of 1kg and has the force 1 * 9.81 = 9.81 N. Dyno 2 carries half the weight of dyno 1 which is 0.50kg and has the force 0.50 * 9.81 = 4.905 N. Is this the correct solution?

As for problem A I am still clueless, I must be a retard. As for the rest, problem E I found this formula: P=F/A but I don't know how to calculate the force in that one. Problem F and G I still don't understand. I am guessing that this is really easy just that my slow brain is having a hard time grasping these formulas and concepts.

 

[phyzguy]

As for problem A, if the blocks are moving at constant speed, what is their acceleration? Given their acceleration, what is the net force on the blocks?

 

[mfb]

Solution: The total mass of the dynos + weights is 1kg, the dyno 1 carries the weight of 1kg and has the force 1 * 9.81 = 9.81 N. Dyno 2 carries half the weight of dyno 1 which is 0.50kg and has the force 0.50 * 9.81 = 4.905 N. Is this the correct solution?

It is correct if the full weight of the dynos is in the lower part. Unfortunately, we don't know this, so the question is a bit unclear.

As for the rest, problem E I found this formula: P=F/A but I don't know how to calculate the force in that one.

That is the right formula for F, not for E.

It is a bit confusing that you use P both for density and pressure.

 

[Fishingaxe]

I want to thank you all for your help in these problems. I appreciate all the help I got. My teacher showed me all the equations for every problem and now I understand it perfectly and won't have any trouble with similar problems in the future. I don't know why but I've always had a hard time with finding the answers for myself but when I see the calculations etc it sticks in my head and I know it for for/to? next time. Must be something really wrong with me. Perhaps it takes practice at thinking on your own to be good at it? or do you think that I am just naturally stupid? I mean I have an A in math and a high IQ(as everybody here has) so I'm not a retard but how come I couldn't figure out these rather simple problems on my own? I am sorry for drifting off topic but I am really interested in hearing the thoughts from intelligent people on this.

 

[mfb]

Perhaps it takes practice at thinking on your own to be good at it?

For sure. Problems are easier to solve if you found solutions for similar problems in the past.