Where W is the object's weight and q is the angle that the incline makes with the horizontal surface.

Where W is the object's weight and q is the angle that the incline makes with the horizontal surface.

To work inclined plane problems:

1. Draw a free body diagram labeling all forces acting on the object. The parallel force is always present and always acts down the incline. The normal force always acts perpendicular to the incline and is used to calculate the magnitude of the frictional force. The frictional force always acts opposite the motion of the object. An applied force is your push or pull that causes the object to move up the incline. We will not have any applied forces acting down the incline.
2. The direction of the motion will always be considered as positive. Forces that act in the direction of the motion are positive and those acting opposite the motion are negative.
3. Find the magnitude of the unbalanced force by adding all the forces together in the direction of the motion (remember to assign signs!).
4. Apply Newton's Second Law to the object's motion. Remember that the unbalanced force is equal to ma. If the object moves at constant speed, ma = 0.

Pulling or Pushing an Object along a Horizontal Surface by Applying a Force at an Angle

When these problems were first introduced, we only considered the case where the object was pushed or pulled along the horizontal surface by a force applied perfectly horizontally to the surface. Now, the force will be applied at an angle q to the horizontal.

To work problems where objects are pushed or pulled at an angle:

1. Draw a free body diagram labeling all forces acting on the object. An applied force is your push or pull that causes the object to slide along the horizontal surface. The applied force is a resultant force. Resolve it into its x- and its y-components.
   • Vertical forces: The y-component acts up if the object is pulled and acts down if the object is pushed. The weight of the object always acts down . The normal force is always present and acts perpendicular to the surface. Assign positive and negative signs to the y-component of the applied force, the normal force, and the weight of the object. Sum the vertical forces together and set equal to zero. This will yield the magnitude of the normal force that is used to calculate the magnitude of the frictional force (if it is not given). The frictional force always acts opposite the motion of the object.
   • Horizontal forces: The x-component of the applied force causes the objects motion. The direction of the motion will always be considered as positive. Forces that act in the direction of the motion are positive and those acting opposite the motion are negative.
2. Find the magnitude of the unbalanced force by adding all the forces together in the direction of the motion (remember to assign signs!).
3. Apply Newton's Second Law to the object's motion. Remember that the unbalanced force is equal to ma. If the object moves at constant speed, ma = 0.

Tension problems:

Tension problems are an example of static equilibrium problems in physics. An object is in stati c equilibrium if the sums of the forces in the vertical direction and in the horizontal direction are zero. The rope can be thought of as a resultant. Resolve it into its x- and y-components. If the rope is attached horizontally, its y-component is zero. If it is attached at an angle, find its components.

To work tension problems:

1. Sketch the x- and y-components for each rope acting at an angle. Ropes pull up! Their y-components act up!!
2. Draw a free body diagram showing all forces that act on the object, horizontal and vertical. If the rope pulls the object to the left, its x-component is negative and its y-component is positive. If the rope pulls the object to the right, its x-component is positive and its y-component is positive. The weight of the object acts purely vertically and is negative.
3. Apply Newton's Second Law. Add all the forces together that act horizontally. Remember signs! Set this sum equal to zero. Algebraically, you may now write an expression for one tension in terms of the other tension.
4. Apply Newton's Second Law. Add all the forces together that act vertically. Remember signs. Set this sum equal to zero. Substitute your expression that you found in step three. You now have one equation in one unknown.

The simultaneous equation solver on the TI-86 can also be used to easily and quickly determine a solution for these problems. The TI-85 simultaneous equation solver operates similarly.

TI-86 Simultaneous Equation Solver

Vector Applications Sample Problems

Vector Applications Homework