Finding static friction of a block on a table on a pulley?

In summary, when the three blocks in Fig. 6-29 are released from rest, they accelerate with a magnitude of 0.500 \frac{m}{s^2}##. Block 1 has mass M, Block 2 has mass 2M, and Block 3 has mass 3M. The coefficient of static friction between Block 2 and the table is 0.37.
  • #1
Eclair_de_XII
1,083
91

Homework Statement


"When the three blocks in Fig. 6-29 are released from rest, they accelerate with a magnitude of ##0.500 \frac{m}{s^2}##. Block 1 has mass M, Block 2 has mass 2M, and Block 3 has mass 3M. What is the coefficient of static friction between Block 2 and the table?"
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Homework Equations


##f_s=(μ_k)(F_N)##
Answer from book: 0.37

The Attempt at a Solution


Okay, so I attempted this by beginning with the sum of all the forces at work, relative to the force on Block 3. There's the gravitational force on Block 1 negating half of Block 3's force, and then there's the static friction between Block 2 and the table also slowing the motion of Block 3. So:

##ΣF = ma = (-9.8\frac{m}{s^2})(M) + (μ_k)(-9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(2M)##
##ΣF = (μ_k)(-9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(M)##
##\frac{ΣF}{m} = a = (9.8\frac{m}{s^2})(-2μ_k + 1)##
##a = \frac{1}{2}\frac{m}{s^2}##
##\frac{1}{2}\frac{m}{s^2} = (9.8\frac{m}{s^2})(-2μ_k + 1)##
##\frac{1}{19.6}=-2μ_k + 1##
##-\frac{18.6}{19.6}=-2μ_k##
##μ_k=\frac{93}{196}##

I have a feeling I messed up on the third step, in calculating acceleration and equating it to ##\frac{1}{2}##. Are there any other forces at work here, that are slowing Block 3's descent? Thank you for anyone who is willing to help me.
 
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  • #2
Your problem statement gives block three a mass of 3M, but it looks like you've used 2M in your first step. Can you verify which is correct?

What mass are you using for 'm' in F = ma? Hint: How much mass is accelerating?
 
  • #3
gneill said:
Can you verify which is correct?

Right. I should be more mindful of reading the problem carefully; especially when doing physics homework.

##ΣF = -(μ_k)(9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(2M)##

gneill said:
How much mass is accelerating?

6M?

##\frac{ΣF}{6M} = -\frac{1}{3}(μ_k)(9.8\frac{m}{s^2}) + \frac{1}{3}(9.8\frac{m}{s^2})##
##\frac{ΣF}{6M} = \frac{1}{3}(9.8\frac{m}{s^2})(1-μ_k) = a##
##\frac{1}{2} = \frac{1}{3}(9.8\frac{m}{s^2})(1-μ_k)##
##\frac{3}{19.6} = 1 - μ_k##
##-μ_k=-\frac{16.6}{19.6}##
##μ_k=\frac{16.6}{19.6}##

I wish I knew what I was doing wrong.
 
Last edited:
  • #4
You say the total amount of mass in the problem is 6M, so then the blocks have masses 1M, 2M, and 3M.

When I do the math assuming those block masses I don't get the book answer. On the other hand, If the third block is 2M and the total mass is 5M, then I get the given answer.

You might want to use a symbol for the gravitational acceleration (g) rather than lug around numbers and units all the way through the algebra. Only plug in numbers at the very end.
 
  • #5
gneill said:
On the other hand, If the third block is 2M and the total mass is 5M, then I get the given answer.

Oops, I should learn to read more carefully.

##ΣF = (-μ_k)(g)(2M) + (g)(M)##
##a = \frac{1}{5}g(1-2μ_k)##
##\frac{1}{2} = \frac{1}{5}g(1-2μ_k)##
##\frac{5}{2g} = 1-2μ_k##
##\frac{5-2g}{2g} = -2μ_k##
##\frac{2g-5}{4g}=μ_k=0.37##

gneill said:
Only plug in numbers at the very end.

Good to note. Thank you...
 

Related to Finding static friction of a block on a table on a pulley?

1. What is static friction?

Static friction is the force that prevents two stationary surfaces from moving against each other.

2. How is static friction different from kinetic friction?

Static friction occurs when two surfaces are not moving relative to each other, while kinetic friction occurs when two surfaces are moving against each other.

3. How do I calculate the static friction of a block on a table on a pulley?

To calculate the static friction, you will need to know the coefficient of static friction between the block and the table, as well as the weight of the block. The formula for static friction is F = μs * N, where F is the force of static friction, μs is the coefficient of static friction, and N is the normal force (equal to the weight of the block in this case).

4. Can the static friction of a block on a table on a pulley change?

Yes, the static friction can change depending on the surface materials, the weight of the block, and the angle of the pulley. It can also change if any external forces are applied to the block.

5. How does the angle of the pulley affect the static friction of the block on the table?

The angle of the pulley can affect the static friction as it can change the normal force acting on the block. The larger the angle, the smaller the normal force and therefore, the smaller the static friction. This is because the normal force is now divided between the vertical and horizontal components.

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