Weight force help...?

Question

I need some help with this problem.
A 215-N box is placed on an inclined plane that makes a 35.0 degree angle with the horizontal. Find the component of the wieght force paralel to the plane's surface.

How do I find it? a clue to which equation I use? and why would the weight force be paralell to the plane's surface?

And note:
I am only looking for help inorder to understand the question more clearly...Do Not Give me the answer!!!

Answer 1

I cannot draw it, then it could be more illustrative and clear.

A mass on an inclined plane with angle (theta) will have gravitational force F = mg with two components : First one parallel to the plane ( which works for movement on the plane) and Second one perpendicular to the plane ( which produces friction agains movement along the plane).

Component parallel to the plane F Sin (theta)
Component perpendicular to the plane F Cos (theta)

So your force component = 215 x Sin (35) N
Please use your calculator and sail

Answer 2

N = W cos(theta) = the normal weight perpendicular to the plane; where W = mg = weight = 215 N and theta = 35 deg.

The weight always acts perpendicular with the horizontal (towards the center of the Earth). This follows because the center of the Earth's mass is in its center and it's the mass (M) of the Earth that causes the gravitational force we call weight.

N, the normal weight, acts perpendicular to the plane; so it acts at an angle with the horizontal when the plane is inclined. We know that N varies as cos(theta) and not sin(theta) because less and less of W would act normal to the plane as the plane's incline increases.

In the end, N = 0 if and only if the planes angle (theta) = 90 deg and none of the box's weight is on the plane and all of the weight is parallel to it. This can only be so if N = W cos(90) = W X 0 = 0. If we mistakenly wrote N = W sin(90), we'd know that was wrong because that would give us N = W X 1 and clearly the weight of a box does not push onto a plane that is inclined 90 degrees straight up from the horizontal.

Thus, using similar logic, the portion of the weight parallel to the plane has to be P = W sin(theta). This follows because P = W sin(90) = W X 1 = W. This means, when the plane is inclined 90 degrees wrt the horizontal, all its weight is parallel to the plane. Conversely, when the incline is theta = 0 deg. (which means there is no incline), then P = W sin(0) = W X 0 = 0; so that there is no parallel force (P) when there is no incline.

The lesson learned: use the extreme angles (0 and 90) to test your equations. If they make sense from a physics point of view, then you are probably right. If they don't, you are probably wrong; so switch the cos for sin, and the sin for cos; and try again with the extremes.

Answer 3

The force due to the weight of the box will be pointing straight down. From the reference point of treating the inclined plane as a level surface, the force will appear to be directed at some angle downward. Any force directed along a plane can be broken up into two components: one parallel to the surface, and one perpendicular to the surface.
An easier way of thinking about it is a right triangle: the hypoteneuse can just be considered a combination of the horizontal and vertical sides of the triangle.
Does that help?

Answer 4

First: Read the ? carefully! It doesnt ask for the weight force, but a COMPONENT of the force. Any force can be broken up into 2 orthoganal vectors and this is commonly done to ease the analysis process.

So draw a little diagram with the plane & weight. Show the weight vector straight down from the weight. Let that vector be the hypoteuse of a right triangle with a short leg going down the incline and another going perpendicular to the incline and joining the tip of the weight vector. The short one is the component of the weight along the plane. (W*sin8) The longer one is the normal force of the weight on the plane (useful for determining friction forces)---(W*cos8)

Assuming you can find the leg lengths of the triangle given that you know the value of the weight...........