Mission F2D1 Vector Components
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A box of mass 'm' is being dragged across a rough, level surface having a coefficient of friction of 'mu.' A force is being applied to the box in an upward and a rightward direction (as shown below). This force makes an angle of 30 degrees with the horizontal. The force of friction experienced by the box would be equal to ...
Students become used to equating the normal force acting upon an object with the weight of the object. After all, the normal force is a support force and supplies the force needed to balance the downward force of gravity. But don't be fooled! The fact is that the normal force is NOT always equal to the object's weight. Questions presented in this mission are perfect examples of the exception to the usual observation.
When there are other vertical forces besides the normal force and the force of gravity, the Fnorm is not equal to the Fgrav. In this question (and others in this mission), there is an applied force that has an upward component. The upward component of the applied force (Fapp-y) can be determined using a trigonometric function (see Math Magic section). The Fnorm value supplies the support needed to make up for the difference between Fgrav and Fapp-y. So for this situation with an upward component of applied force, the Fnorm value is Fgrav - Fapp-y. The force of friction can be determined using the equation presented in the Formula Frenzy section and the value of Fnorm.
Ffrict= mu • Fnorm
where mu is the coefficient of friction (dependent predominantly upon the nature of the two surfaces that are in contact) and Fnorm is the normal force.
