Maximizing Friction Coefficient for Block on a Wedge: A Calculus Approach

In summary, the conversation discusses a problem involving a block of mass m on a rough wedge inclined at an angle α to the horizontal. The coefficient of kinetic friction between the block and wedge is given by µ_0x/d, where x is the distance down the slope from the starting point. The goal is to calculate the maximum value of µ_0 that will allow the block to reach the bottom of the wedge. The attempt at a solution involves drawing a diagram and resolving the forces, which leads to finding the acceleration of the block in terms of x. However, the acceleration is defined over x rather than time, making it difficult to proceed. The use of the chain rule is suggested, but it is unclear how x depends on time. Further
  • #1
Silicon-Based
51
1

Homework Statement


A block of mass [itex]m[/itex] is placed on a rough wedge inclined at an angle [itex]α[/itex] to the horizontal, a distance [itex]d[/itex] up the slope from the bottom of the wedge. The coefficient of kinetic friction between the block and wedge is given by [itex]µ_0x/d[/itex], where [itex]x[/itex] is the distance down the slope from the starting point. Calculate the maximum value of [itex]µ_0[/itex] which will allow the block to reach the bottom of the wedge.

2. The attempt at a solution
After drawing a diagram and resolving the forces I found the acceleration of the block in terms of [itex]x[/itex]:
$$a(x)=g(\sin(\theta)-x\frac{µ_0}{d}\cos(\theta))$$
I don't know how to proceed further, given that the acceleration is defined over [itex]x[/itex] rather than [itex]t[/itex], which prevents me from simply integrating this expression. I suspect the chain rule could be useful ([itex]a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}[/itex]), but here again, I don't know how [itex]x[/itex] depends on [itex]t[/itex].
 
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  • #2
Silicon-Based said:

Homework Statement


A block of mass [itex]m[/itex] is placed on a rough wedge inclined at an angle [itex]α[/itex] to the horizontal, a distance [itex]d[/itex] up the slope from the bottom of the wedge. The coefficient of kinetic friction between the block and wedge is given by [itex]µ_0x/d[/itex], where [itex]x[/itex] is the distance down the slope from the starting point. Calculate the maximum value of [itex]µ_0[/itex] which will allow the block to reach the bottom of the wedge.

2. The attempt at a solution
After drawing a diagram and resolving the forces I found the acceleration of the block in terms of [itex]x[/itex]:
$$a(x)=g(\sin(\theta)-x\frac{µ_0}{d}\cos(\theta))$$
I don't know how to proceed further, given that the acceleration is defined over [itex]x[/itex] rather than [itex]t[/itex], which prevents me from simply integrating this expression. I suspect the chain rule could be useful ([itex]a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}[/itex]), but here again, I don't know how [itex]x[/itex] depends on [itex]t[/itex].
If you rewrite the equation as ##\ddot x+Bx+C=0##, does it remind you of anything?
Alternatively, use that chain rule, but remember that dx/dt is v.
 
  • #3
Update:

If I were to integrate the acceleration with respect to time I would get:
$$v(x,t)=g(\sin(\theta)-x\frac{\mu_0}{d}\cos(\theta))t$$ Knowing that the maximum occurs when [itex]v(x=d)=0[/itex], this equation gives [itex]\mu_0=\tan(\theta)[/itex].

Alternatively, the chain rule gives [itex]a=\frac{dv}{dx}v[/itex], so $$\int a dx=\int v dv$$ Again, applying the above condition, I end up with [itex]\mu_0=2g\tan(\theta)[/itex]. Which one is correct?
 
  • #4
Silicon-Based said:
Update:

If I were to integrate the acceleration with respect to time I would get:
$$v(x,t)=g(\sin(\theta)-x\frac{\mu_0}{d}\cos(\theta))t$$ Knowing that the maximum occurs when [itex]v(x=d)=0[/itex], this equation gives [itex]\mu_0=\tan(\theta)[/itex].

Alternatively, the chain rule gives [itex]a=\frac{dv}{dx}v[/itex], so $$\int a dx=\int v dv$$ Again, applying the above condition, I end up with [itex]\mu_0=2g\tan(\theta)[/itex]. Which one is correct?
Neither.
The first is wrong because you cannot integrate x wrt t merely by multiplying it by t. x depends on t.
The second cannot be right because it is dimensionally wrong. You have a dimensionless number on the left and an acceleration on the right. But the method looks ok, so please post your working.
 

Related to Maximizing Friction Coefficient for Block on a Wedge: A Calculus Approach

1. What is a block on a wedge?

A block on a wedge is a common physics problem where a block is placed on a wedge-shaped surface and is subjected to a gravitational force acting downward.

2. What is the purpose of studying block on a wedge?

Studying block on a wedge allows us to understand the concepts of equilibrium, forces, and motion in a practical and visual way. It also helps us to apply mathematical concepts, such as calculus, to solve real-world problems.

3. How do you calculate the normal force on a block on a wedge?

The normal force can be calculated by using the formula Fn = mg cosθ, where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the wedge.

4. What is the relationship between the angle of the wedge and the friction force?

The angle of the wedge and the friction force have an inverse relationship. As the angle of the wedge increases, the friction force decreases. This is because a steeper angle results in a smaller area of contact between the block and the wedge, reducing the force of friction.

5. How can calculus be used to solve problems involving block on a wedge?

Calculus can be used to analyze the forces and motion of a block on a wedge by calculating the derivatives and integrals of the forces involved. This allows us to determine the acceleration, velocity, and position of the block at any given time.

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