Angled Forces: Finding Components, Normal Force, and Acceleration

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To solve the problem of a 57.4 kg crate on a 30-degree incline, the components of the weight forces must be calculated: the parallel force (Fgx) and the perpendicular force (Fgy). The normal force can be determined using the perpendicular component of the weight. Acceleration down the plane can be calculated using the forces acting on the crate, and a different weight of 5620 N will yield a different acceleration. It is established that acceleration does depend on weight but is independent of mass in this context.
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Angled Forces HELP!

QUESTION: A 57.4 kg crate is resting on a plane inclined 30 degrees above the horizontal.
a) Find the compoents of weight forces that are parallel to the plane (Fgx) and perpendicular to the plane (Fgy)

b) Find the normal force

c) Find the acceleration down the plane

d) Find the accleration down the plane if the crate weighed 5620 N

e) does acceleration depend on weight? Does it depend on mass?



2. Homework Equations - How do you set up this problem, or do anything? I'm so lost and I could really use a step by step explanation.



3. An attempt at a solution:

I drew a diagonal line with a box on it. Then I have my Normal force going directly perpendicular to the box, and my Fg going straight down from the box... Then I got lost :cry:
 
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Mbenjj0x0xx said:
QUESTION: A 57.4 kg crate is resting on a plane inclined 30 degrees above the horizontal.
a) Find the compoents of weight forces that are parallel to the plane (Fgx) and perpendicular to the plane (Fgy)

b) Find the normal force

c) Find the acceleration down the plane

d) Find the accleration down the plane if the crate weighed 5620 N

e) does acceleration depend on weight? Does it depend on mass?



2. Homework Equations - How do you set up this problem, or do anything? I'm so lost and I could really use a step by step explanation.



3. An attempt at a solution:

I drew a diagonal line with a box on it. Then I have my Normal force going directly perpendicular to the box, and my Fg going straight down from the box... Then I got lost :cry:


Start by drawing the 57.4 kg force vector straight down from the incline. From the tip of that vector draw a line perpendicularly back up to the lncline. That forms a little 30-60-90 triangle with your normal force vector as the hypotenuse. The length of the other legs of that triangle give the magnitudes of the normal and parallel forces. You can figure those out because you know the angles.
 
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