When an object is sliding down or resting on an incline it is often convenient t

Question

When an object is sliding down or resting on an incline it is often convenient to think about axes that are parallel and perpendicular to the incline rather than horizontal and vertical. The figure below shows a box resting on an incline.

In this situation the x-axis is defined to be parallel to the incline and positive up and to the right. The y-axis is perpendicular to the surface. The challenge is that the force due to gravity is still straight down, so you have to break the force to due to gravity down into a component that is parallel to the surface and a component that is perpendicular to the surface. These components are shown in red on the diagram above (the x-component is shown twice, once on the axis, and once moved over to help complete ?the triangle? you will need to use when find the components).
Suppose the box weighs 32.0 N and the angle of the incline is ?=28.2?

A.) For this incline, what is the x-component of the force due to gravity? This would be the component of the weight which is trying to pull the box down the incline to the left. In this case include a positive or negative sign which is consistent with the axes chosen in the picture.

B.) What is the y-component of the force due to gravity? (This would be the component of the weight which is trying to hold the box onto the incline) In this case include a positive or negative sign which is consistent with the axes chosen in the picture.

Explanation / Answer

A)

Fx = Fg*sin theta = 32*sin28.2 = 15.12 N

B)

Fy = -Fg*cdos theta = -32*cos28.2 = 28.2 N

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