# Homework Help: Force, Mass and Acceleration Problem

1. Nov 21, 2012

### blondii

1. The problem statement, all variables and given/known data
A block of mass m1 on a rough horizontal surface is connected to a ball of mass m2 by a light weight cord over a light weight, friction-less pulley, as shown in the figure below. A force of magnitude F at an angle θ with the horizontal is applied to the block as shown. The coefficient of kinetic friction between the block and surface is uk. Determine the magnitude of the acceleration of the two objects.

2. Relevant equations
Refer to attachment

3. The attempt at a solution
I would just like to confirm my answer. Better still if anyone can suggest a shorter way of solving it would be nice. Please find my complete solutions on pdf attached. Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### Physics.pdf
File size:
858.1 KB
Views:
158
2. Nov 21, 2012

### Staff: Mentor

We can't help yet until you show us what you've done.

Is that your answers written in the Pdf and hilighted in turquoise? and is that your work shown above the hilighting?

It looked like this was a PDF question/answer from the book?

3. Nov 21, 2012

### haruspex

If you look at the legend at top left of page 2 you can see it is not out of a book. More tellingly, it's wrong.
blondii, your force diagram for the block omits the tension in the string. Also, using F for the string tension in the second free body diagram creates confusion. Put, say, T for the tension in both diagrams and try again.

4. Nov 22, 2012

### blondii

I have reworked the problem with tensions added as specified. See my new attached diagram. Here is my new working out also. Please confirm if my answer is correct this time or any other suggested approaches. Thanks:

Let:
fk = Kinetic Friction
uk = Coefficient of Kinetic friction
g = Gravitational Acceleration
n = Normal
T = Tension

Force acting on block
ƩFx = Fcos - fk - T = m1a
Eq (1)
∴ Fcos - fk - T = m1a

ƩFy = Fsinθ + n - w1 = 0
∴ n = w1 - Fsinθ
Eq (2)
= m1g - Fsinθ (2)

Substitute (2) into (1)
Fcosθ - uk(m1g - Fsinθ) - m1a
Eq (A)
∴T = Fcosθ - uk(m1g - Fsinθ) - m1a

Force acting on ball
ƩFy = T -w2
= T - m2g = m2a
Eq (B)
∴ T = m2a + m2g

Calculate Acceleration
Substitute (B) into (A)
m2g + m2a = Fcosθ - uk(m1g-Fsinθ) - m1a
a(m1 + m2) = Fcosθ - uk(m1g - Fsinθ) - m2g
∴ a = (Fcosθ - uk(m1g - Fsinθ) - m2g) / (m1 + m2)

#### Attached Files:

• ###### Physics Diag 2.jpg
File size:
19 KB
Views:
193
5. Nov 22, 2012

### blondii

Please find attached also my revised answer on PDF and confirm if I am on the right track now. Thanks

#### Attached Files:

• ###### Physics.pdf
File size:
906.6 KB
Views:
127
6. Nov 22, 2012

### grzz

7. Nov 22, 2012

### blondii

Thanks for confirming grzz. Much appreciated

8. Nov 22, 2012

### Staff: Mentor

how did you do the drawings? they look very professional.

9. Nov 22, 2012

### blondii

I did the drawings using Adobe Illustrator

10. Nov 22, 2012

### Staff: Mentor

Okay, thanks.