Static Friction: Backpack & Table in Physics Lab

AI Thread 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.
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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
 
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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.
 
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