Static Friction: Backpack & Table in Physics Lab

Click For Summary
The discussion revolves around calculating the coefficient of static friction between a backpack and a table using physics principles. A backpack weighing 47.0 N is attached to a spring with a force constant of 150 N/m, which stretches 2.20 cm before the backpack begins to slip. The maximum frictional force is determined by the equation Fsmax = μs * FN, where FN is the normal force equal to the weight of the backpack. By equating the spring force (3.3 N) to the maximum frictional force, the coefficient of static friction is calculated to be approximately 0.0702. The conversation emphasizes the importance of understanding the forces involved and using free-body diagrams for clarity.
stealth2k4
Messages
2
Reaction score
0
A backpack full of books weighing 47.0 N rests on a table in a physics laboratory classroom. A spring with a force constant of 150 N/m is attached to the backpack and pulled horizontally. If the spring stretches by 2.20 cm before the backpack begins to slip, what is the coefficient of static friction between the backpack and the table?

i tried to use Fsmax = Mus of s times the normal force what should i use
 
Last edited:
Physics news on Phys.org
You have the equation. What is Fsmax? What is the normal force?

Use a free-body diagram to see all the forces.
 
stealth2k4 said:
A backpack full of books weighing 47.0 N rests on a table in a physics laboratory classroom. A spring with a force constant of 150 N/m is attached to the backpack and pulled horizontally. If the spring stretches by 2.20 cm before the backpack begins to slip, what is the coefficient of static friction between the backpack and the table?

i tried to use Fsmax = Mus of s times the normal force what should i use
{Normal Force of Books on Table} = FN = (47.0 N)
{Coefficient of Static Friction} = K
{Max Frictional Force on Books} = K*FN = (47)*K
{Spring Force Constant} = c = (150 N/m)
{Spring Displacement} = d = (2.20 cm) = (2.2e(-2) m)
{Spring Force Applied to Books} = c*d = (150 N/m)*(2.2e(-2) m) = (3.3 N)

Since books began to move with the above spring displacement, we have:
{Spring Force Applied to Books} = {Max Frictional Force on Books}
::: ⇒ (3.3) = (47)*K
::: ⇒ K = (0.0702)


~~
 
xanthym said:
{Normal Force of Books on Table} = FN = (47.0 N)
{Coefficient of Static Friction} = K
{Max Frictional Force on Books} = K*FN = (47)*K
{Spring Force Constant} = c = (150 N/m)
{Spring Displacement} = d = (2.20 cm) = (2.2e(-2) m)
{Spring Force Applied to Books} = c*d = (150 N/m)*(2.2e(-2) m) = (3.3 N)

Since books began to move with the above spring displacement, we have:
{Spring Force Applied to Books} = {Max Frictional Force on Books}
::: ⇒ (3.3) = (47)*K
::: ⇒ K = (0.0702)


~~

Please don't solve the entire problems. They won't learn from that.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
1
Views
7K
Replies
13
Views
3K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 1 ·
Replies
1
Views
4K