Force, Mass and Acceleration Problem

[blondii]

Homework Statement

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 u_k. Determine the magnitude of the acceleration of the two objects.

Homework Equations

Refer to attachment

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

 

[jedishrfu]

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?

 

[haruspex]

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

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.

 

[blondii]

Thanks for your response haruspex.

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:
f_k = Kinetic Friction
u_k = Coefficient of Kinetic friction
g = Gravitational Acceleration
n = Normal
T = Tension


Force acting on block
ƩF_x = Fcos - f_k - T = m₁a
Eq (1)
∴ Fcos - f_k - T = m₁a

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

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

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

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

 

[blondii]

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.

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

 

[grzz]

Your final answer is ok.

 

[blondii]

Your final answer is ok.

Thanks for confirming grzz. Much appreciated

 

[jedishrfu]

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

 

[blondii]

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

I did the drawings using Adobe Illustrator

 

[jedishrfu]

I did the drawings using Adobe Illustrator

Okay, thanks.