Finding acceleration of block with block on top, connected by rope

[sentimentaltrooper]

Homework Statement

A rope pulls on the lower block in the figure with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.16. The coefficient of kinetic friction between the lower block and the upper block is also 0.16. The pulley has no appreciable mass or friction. What is the acceleration of the 2.0 kg block?

Relevant Equations

F=ma
a=F/m

I have already concluded that the way to solve this problem is through
(20 N - (3 kg * 9.8 m/s^2 * 0.16) - (2*(1 kg * 9.8 m/s^2 * 0.16))) / 3 kg

I have several questions:

• Why do we multiply the second set of parentheses by 2? Why do we count the friction between the 1 kg box and the 2 kg box twice?
• Also, the fact that the frictions are both 0.16 does not allow me to differentiate between the calculations. The first set of parentheses with 3 * 9.8 * 0.16 is multiplied by 0.16 because that is the friction between the lower block and the surface correct?

Thank you

 

[Delta2]

This problem seems "wicked" (at least to me) but can be solved relatively easily with repetitive applications of Newton's 2nd law in the horizontal and vertical direction. Both of your questions will be answered if you correctly and repeatedly apply Newton's 2nd law and combine correctly the resulting equations.

 

[haruspex]

Why do we count the friction between the 1 kg box and the 2 kg box twice?

One way to think of it is that the relative motion of the two surfaces is double that of the relative motion of the lower block and the ground. So for a given frictional force, twice as much work is done there for each unit of distance by which the 20N force advances.

 

[sentimentaltrooper]

I thank you for your responses and I apologize for this but for some reason I still cannot understand why it is double, or twice as much work.

 

[haruspex]

I thank you for your responses and I apologize for this but for some reason I still cannot understand why it is double, or twice as much work.

If the applied 20N force advances distance x, the lower block moves x relative to the ground. If the the frictional force there is $F_L$ then the work done is $xF_L$. At the same time, the upper block moves x the other way, so the surfaces of the two blocks in contact move 2x wrt each other. If the frictional force there is $F_U$ then the work done against that is $2xF_U$.

 

[sentimentaltrooper]

If the applied 20N force advances distance x, the lower block moves x relative to the ground. If the the frictional force there is $F_L$ then the work done is $xF_L$. At the same time, the upper block moves x the other way, so the surfaces of the two blocks in contact move 2x wrt each other. If the frictional force there is $F_U$ then the work done against that is $2xF_U$.

Thank you so much for this.

 

[anuttarasammyak]

Say 2kg body has acceleration $a$, 1 kg body has acceleration $-a$ due to binding by rope. We can make two equations of motion for the two bodies and get $a$ from them.