Forces of friction with angles

AI Thread Summary
To solve friction problems involving angles, it's essential to break down the applied force into its horizontal and vertical components using trigonometric functions. In the case of a 1000-N crate pushed by a 300-N force at a 20-degree angle below the horizontal, the horizontal component is calculated as Fcos(20), while the vertical component is Fsin(20). The coefficient of kinetic friction can then be determined using the formula μ = (Fcos(20)) / (m*g), where m is the mass derived from the weight of the crate and g is the acceleration due to gravity. After substituting the values, the coefficient of kinetic friction is approximately 0.061. Understanding these components is crucial for accurately calculating friction in angled scenarios.
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I know the formula for friction coefficients is force over weight, but i don't understand what to do when angles are involved. Please help me by showing me where to put the angle measurements in this problem.

A 1000-N crate is being pushed across a level floor at a constant speed by a force F of 300N at an angle of 20 degrees below the horizontal. What is the coefficent of kinetic friction between the crate and the floor?
 
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your going to use vector addition. you must find the force in its components. (x and y). You can use this using simple trig laws. it helps to draw a free body diagram. good luck!
 


When angles are involved in friction problems, we need to consider the components of the forces involved. In this case, we have a force F of 300N at an angle of 20 degrees below the horizontal, and the weight of the crate, which is 1000N, acting vertically downwards.

To find the coefficient of kinetic friction, we can use the formula: μ = F/ (m*g), where μ is the coefficient of friction, F is the force applied, m is the mass of the crate, and g is the acceleration due to gravity (9.8 m/s^2).

However, since the force F is acting at an angle, we need to find the horizontal component of this force, which is Fcos(20). This is the force that is actually moving the crate horizontally. The vertical component, Fsin(20), is counteracted by the weight of the crate.

Now, we can plug in the values into the formula: μ = (Fcos(20))/ (m*g). Since we are given the force F as 300N and the weight of the crate as 1000N, we can rewrite the formula as: μ = (300cos(20))/ (1000*9.8).

Solving this equation gives us a coefficient of kinetic friction of approximately 0.061. This means that for every 1N of horizontal force applied, there is a frictional force of 0.061N acting in the opposite direction.

In summary, when angles are involved in friction problems, we need to consider the horizontal and vertical components of the forces involved and use the appropriate components in the formula for coefficient of friction.
 
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