DYNAMICS

Dynamics



Dynamics

Question 1

Draw free-body diagrams for each of the objects, including all forces acting on them;

(b) Solve for the conditions (in terms of the ratios mb/ma and/or mc/ma) required in order for the block A to be raised up the incline

Let the tension in the cord connecting B and C be Tbc, and the tension in the cord connecting B and A be Tba. Mass C has only two forces acting on it: Tbc and its weight mcg. Clearly the acceleration is downward, in the same direction as the weight and opposite the tension.

Tbc - mcg = -mca

Mass A has only one force acting on it, the tension Tab, giving

Tab = maa

This is not quite enough information. However, since B and C are connected together, we may treat them, from the point of view of the upper cord, as a single mass (mb + mc) connected to mass A. There are two forces acting on B and C connected together: their weight, and the tension

Tab. Thus,

Tab - (mb + mc) g = - (mb + mc) a (23)

Since we already know Tab =maa

mcma + mcmb + mc 2- mcmb - mc2]

ma+mb +me

maa - (mb + mc) g = - (mb + mc) a



(c) Determine the acceleration of block A if it is being raised up the incline;

Answer: Only that component of the applied force which is parallel to the incline has any influence on the block's motion: the normal component of the applied force is canceled out by the normal reaction of the incline. The component of the applied force acting up the incline is . Likewise, the component of the block's weight acting down the incline is. Hence, using Newton's second law to determine the acceleration  of the block up the incline, we obtain 

Since  and , we have 

(d) Determine ...