Kinetic friction and things

[Lmbrando]

Hi I need help on my homework and i have no idea where to start and my teacher gives us webassign problems but he does not teach us how to do them, he just gives us formulas.. If anyone can help me with at least this problem i'd appreciate it thx

p.s I am struggling help!



A 1400 N crate is being pushed across a level floor at a constant speed by a force of 300 N at an angle of 20.0° below the horizontal, as shown in Figure P4.37a.



(a) What is the coefficient of kinetic friction between the crate and the floor?

(b) If the 300 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in Figure P4.37b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).
m/s2

 

[Hootenanny]

Hi there Lmbrando and welcome to PF,

Have you any thoughts on the matter thus far?

 

[kyle8921]

Sure, I can help you with this problem! To start, let's define some terms and equations that will be helpful in solving this problem.

Kinetic friction is the force that opposes the motion of an object as it slides against another surface. It is caused by the microscopic roughness of the surfaces in contact, which creates resistance to movement.

The coefficient of kinetic friction, denoted by μ, is a dimensionless quantity that represents the ratio of the force of kinetic friction to the normal force between the two surfaces. It is a property of the materials in contact and is often listed in tables or can be experimentally determined.

To solve this problem, we will use Newton's second law, which states that the sum of all forces acting on an object is equal to its mass times its acceleration. In equation form, this is written as ΣF = ma.

Now, let's apply this to the given scenario. The crate has a weight of 1400 N, which means that the force of gravity acting on it is 1400 N downwards. The 300 N force pushing or pulling the crate is also acting on it, but at an angle. To find the net force, we need to break this force into its horizontal and vertical components. Using trigonometry, we can find that the horizontal component is 300*cos(20°) = 283.4 N and the vertical component is 300*sin(20°) = 103.1 N.

(a) To find the coefficient of kinetic friction, we need to first find the normal force between the crate and the floor. This is the force that the floor exerts on the crate to support its weight. Since the crate is not accelerating, the normal force must be equal and opposite to the weight of the crate. Therefore, the normal force is also 1400 N.

Now, we can use Newton's second law to find the force of kinetic friction. Since the crate is moving at a constant speed, the net force on it must be zero. This means that the force of kinetic friction must be equal and opposite to the horizontal component of the 300 N force. Therefore, we can set up the following equation:

μN = 283.4 N

Solving for μ, we get:

μ = 283.4 N / 1400 N = 0.2025

Therefore, the coefficient of kinetic friction between the crate and the floor is 0.202