Solve Friction on an Angle with 7.0kg Mass: u=0.2

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SUMMARY

The discussion focuses on calculating the acceleration of a 7.0 kg mass on a 42-degree ramp with and without friction. When friction is absent, the acceleration is determined to be 6.6 m/s², calculated using the gravitational force component along the ramp. With a coefficient of friction of 0.2, the new acceleration must be recalculated by considering the frictional force, which is derived from the normal force and the coefficient of friction.

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  • Knowledge of trigonometric functions (sine, cosine) in right triangles
  • Familiarity with free body diagrams and vector decomposition
  • Basic principles of friction and normal force calculations
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  • Calculate the new acceleration down the ramp using the formula: a = g sin(θ) - (μ * g cos(θ))
  • Explore the effects of varying the angle of inclination on acceleration and friction
  • Study the relationship between mass, friction, and acceleration in inclined planes
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A 7.0 Kg mass is sitting on a ramp that is angled 42 Degreees above the horizontal.
a) Find the value of the acceleration the mass experiences going down the ramp if there is no friction between the mass and the ramp.
b) Determine the acceleration down the ramp if the coefficient of friction between the mass and the ramp is 0.2.




Equations being used in this question.
Ff = u Fn
Fn = mg



Attempt at solution
A) I cannot draw triangles on here. I have a right angle triangle with the right angle in the lower right side. The right side is 9.8m/s^2, the opposite angle is 42 degrees.

so triangle looks like /|basically.

Using Soh Cah Toa, I find that the Hypotenuse of said Triangle is:
sin42 = 9.8/H (H being the hypotenuse)
H = 9.8/sin42
H = 14.65 m/s^2

However, according to my sheet the answer is: 6.6m/s^2

B) I do not know how to go about doing this. If the acceleration going down the ramp is 6.6m/s^2, and the coefficient of friction is 0.2, how do I find the new acceleration?
 
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In your free body diagram, the weight, which acts vertically downwards, can be split into two components, one is perpendicular to the slope and the other is parallel to the slope. The angle between the vertically downard weight and the component perpendicular to the slope is the same as angle of inclination of the slope.Can you form a vector triangle for the weight now?


For a normal reaction,N, and coefficient of friction,[itex]\mu[/itex], the frictional force,FR is given by [itex]F_R= \mu N[/itex].
 

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