A 5 kg cart rests on an incline which makes an angle of 30 degrees with the horizontal.
Sketch this situation with the incline rising as you move to the right and the cart on the incline. Include an x-y coordinate system with the origin centered on the cart, with the x axis directed up and to the right in the direction parallel to the incline.
The gravitational force on the cart acts vertically downward, and therefore has nonzero components parallel and perpendicular to the incline.
Sketch the x and y components of the force, as estimate the magnitude of each component.
What angle does the gravitational force make with the positive x axis, as measured counterclockwise from the positive x axis? Which is greater in magnitude, the x or the y component of the gravitational force?
The magnitude of the gravitational force is 5 kg * 9.8 m/s^2 = 49 N, which should be indicated in your sketched.
Sketched in the given orientation, the x axis is directed 30 degrees above horizontal. The gravitational force is directed vertically downward, which puts it at 90 degrees relative to horizontal. To get from the positive x axis to the gravitational force vector therefore requires a counterclockwise turn through 240 degrees (or a clockwise turn through 120 degrees). The requested angle is counterclockwise, therefore it is 240 degrees.
This makes the direction of the gravitational force closer to the y axis than to the x axis, so that its y component has a greater magnitude than its x component.
(note also that the x and y components of this force are both negative)
Using the definitions of the sine and cosine, find are the components of the cart's weight parallel and perpendicular to the incline.
The x component is magnitude * cos(theta) = 49 N * cos(240 degrees) = -24.5 N.
The y component is magnitude * sin(theta) = 49 N * sin(240 degrees) = -43 N, approx.
NOTE: The text gives formulas based on right-angle trigonometry, which are valid and with proper documentation may be used. However the picture given here is recommended. The text's formulas use theta as the angle of inclination of the incline, and reverse the roles of the sine and cosine formulas. Once you understand the picture presented here, you may make a decision about which approach works better for you; however if you aren't well-versed in right-angle trigonometry, you should use the picture described here.
How much elastic or compressive force must the incline exert to support the cart, and what is the direction of this force?
The y component of the gravitational force is -43 N. The y axis is perpendicular to the incline, so this -43 N force component acts perpendicular to the incline. The incline responds with an elastic or compressive for of +43 N, in the positive y direction. This force is therefore perpendicular to the incline, directed upward and to the left.
Note that the x axis is parallel to the incline. The x component of the gravitational force is -24.5 N, parallel to the incline and acting in the downward direction. This force is not affected forces in the y direction, and in the absence of friction or other forces acting parallel to the incline this force meets no opposition, causing the cart to accelerated down the incline.
If no other force is exerted parallel to the incline, what will be the cart's acceleration?
The net force on the cart will be -24.5 N in the x direction--i.e., 24.5 N in the direction down the incline.
The corresponding acceleration will be a = F_net / m = -24.5 N / (5 kg), in the x direction. This is -4.9 m/s^2 in the x direction, i.e., 4.9 m/s^2 down the incline.