Kinetic Friction with Blocks Need Help.

[Larin]

Homework Statement

Suppose the coefficient of kinetic friction between m₁ and the plane in the figure is μ_k=0.20, and that m₁=m₂=2.2kg.
a) As moves down, determine the magnitude of the acceleration of m₁ and m₂ given θ=28^o.

b) What smallest value of μ_k will keep this system from accelerating?

Homework Equations

F=ma

The Attempt at a Solution

I am not sure even where to begin.

 

[pgardn]

So what are the forces acting on both masses? Can you draw FBD's to help you get some equations to solve for a?

 

[Larin]

The normal force, the force of the weight, and the force of tension are acting on both blocks.

 

[Larin]

I am still really confused though...

 

[asteriatic]

I can provide some guidance on how to approach this problem. First, we need to understand the concept of kinetic friction. Kinetic friction is the force that opposes the motion of an object when it is in motion. In this case, the blocks are moving down the plane, so there will be a force that opposes this motion.

To solve for the acceleration of the blocks, we can use the equation F=ma. We know that the force of friction, Ff, is equal to the coefficient of kinetic friction, μk, multiplied by the normal force, Fn. The normal force is the force that the plane exerts on the blocks perpendicular to the surface. So, we can rewrite the equation as Ff=μkFn.

Since the blocks are moving down the plane, the force of gravity, Fg, will also be acting on them. This force can be broken down into its components along the plane and perpendicular to the plane. The component perpendicular to the plane will be equal to mgcosθ, where m is the mass of the blocks and g is the acceleration due to gravity. This component will be balanced by the normal force, so we can write Fn=mgcosθ.

Now, we can substitute this value for Fn into our equation for friction, giving us Ff=μk(mgcosθ). We know that this force is equal to the mass of the blocks multiplied by their acceleration, so we can write ma=μk(mgcosθ). Solving for a, we get a=μkgcosθ/m.

For part a) of the problem, we are given the mass of the blocks and the value of θ. Plugging these values into our equation, we get a=0.20(2.2kg)(9.8m/s^2)cos28o/2.2kg. This gives us an acceleration of approximately 1.56 m/s^2 for both blocks.

For part b) of the problem, we need to determine the smallest value of μk that would prevent the blocks from accelerating. In this case, we want the force of friction to be equal and opposite to the component of gravity along the plane, so that the net force on the blocks is zero. This means that we want Ff=μk(mgcosθ)=mgcosθ. Solving for μk, we get μk=cosθ. Plugging in the given value