# Homework Help: Basic Mechanical Engineering Help Required

1. Aug 16, 2011

### sperrya

Hi there, hoping someone can help me. I am studying ICS distance learning course and, well to be honest am getting no support whatsoever from the course tutor AND have discovered lots of mistakes in their learning modules. I have no faith in them now whatsoever! However, I need to complete the following questions, but am struggling... ANY help to explain how I answer the following questions will be much appreciated! I have attached a picture of the initial problem.

1. The problem statement, all variables and given/known data

Q1. Calculate the component of the force that is acting horizontally on the crate
Q2. State the effect the force has on the line connecting the crate to the motor
Q3. Each crate has a mass of 200kg, calculate the acceleration experienced by each crate
Q4. The distance covered by each crate is 20m. Assuming that each crate starts at rest, calculate the speed of each crate at the end of the 20m journey
Q5. Calculate the kinetic energy of the crate at the end of the journey
Q6. Calculate the duration of the journey
Q7. Calculate the work done on the crate during the journey
Q8. Calculate the change in momentum of each crate

2. Relevant equations

See below.

3. The attempt at a solution

I have attempted some of the questions as follows:

Q1. 500Cos15 = 482.96N
Q2. The force increases the tension on the line connecting the crate to the motor and pulls the crate across the floor
Q3. a=F/M so a=500/200, therefore a=2.5ms-2
Q4.
Q5.
Q6.
Q7. WD=F x S, so 482.96 x 20 = 9659.2 Joules
Q8.

Any help would be much appreciated, especially Q4..

Adam (new to mechanical engineering and learning!)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### untitled.png
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2. Aug 16, 2011

### PeterO

Q3. the horizontal component [Q1] is causing the acceleration not the full 500N

Q4 - Q8 you should be applying the usual equations of motion, Newtons laws and expressions for Kinetic Energy and Momentum

3. Aug 16, 2011

### sperrya

So, does this look right..??

Q1. 500Cos15 = 482.96N
Q2. The force increases the tension on the line connecting the crate to the motor and pulls the crate across the floor
Q3. a=F/M so a=482.96/200, therefore a=2.415ms-2
Q4. v=2s/t, therefore v=2.42ms-1
Q5. KE=1/2MV2, therefore KE=585.64 Joules
Q6. V=U+at, therefore duration is 16.56 seconds
Q7. WD=F x S, so 482.96 x 20 = 9659.2 Joules
Q8. Momentum=MV, therefore 200 x 2.415 = 483kgms-1

Your help/guidance would be much appreciated. The material ICS supplied has been teaching me the wrong principles so forgive me if I seem confused over what seems easy questions.

Thanks

4. May 6, 2012

### mechnewbie

i beg to differ.....

Q1. 482.96N
Q2. same
Q3.2.415ms-2

Q4.... i have a different answer to sperrya's....

i used the formula V²= U² + 2as (in the course notes)
which results in an answer of 9.818m-1

You're not the only one who is confused!!

Last edited: May 6, 2012
5. Jun 13, 2012

### cps.13

I, unfortunately, am doing the same course! I wish I had done more research into ICS course material first, full of mistakes and missing info.

I have different answers from Q4 on...

Q4 - 9.82 ms-1
Q5 - 9.64 kJ
Q6 - 4.07 seconds
Q7 - 9659.20 J
Q8 - 1,964 kg ms-1
Q9 - 6.547 kN

6. Jun 20, 2012