The addition of forces in dynamics

December 26th, 2011

The dynamics equations are based on the determination of the causes of body movements and their impact on the type of movement. The dynamic equations are the solution to the basic problems of mechanics, but these solutions are expressed using dynamic variables: force, energy, momentum.  

  With the differential relations, we can obtain the dynamic equations from the kinematics and with using integral relations can be obtained from the kinematic equations the equations of dynamics.

    The solution of fundamental problems of mechanics in dynamics is based on the elucidation of the causes that lead to an appearance of  a movement. These reasons are the interactions between bodies. A quantitative measure of these interactions is a force.

      The force is a vector quantity, which is a quantitative measure of interaction of bodies and has three characteristics: the direction, the module, the point of application. The presence of the force application point distinguishes it from other vector quantities in physics and leads to a number of features associated with the addition of forces.

     The first feature is that the force application point can be moved in the direction of the force:  the result of the action of the force will not change. In fact, the locomotive can push a railway carriage or pull it on a trailer: the result of this action will not change (animation 1, 2).

Animation 1

Animation 2

The second feature, which is a consequence of the first, is that the transfer of the points of application of force in the direction of their actions can adduce the forces acting on the body, to the total top and then add the strengths according to the rules of the addition of vectors.

 Let’s consider an example in which two students trying to lift a heavy suitcase by applying two forces \vec F_1 (Fig.1A).

addit_forces

Fig.1. The addition of forces.

     Carrying out a gradual transfer of the application points of forces and, as shown in 1B, performs addition of forces  \vec F_1 и \vec F_2, which are reduced to a common origin, as shown in Fig.1C, and replace the general action of forces \vec F_1 и \vec F_2 by the action of the force F=\vec F_1+\vec F_2, which applies to the suitcase porter.

    The force \vec F makes the action equal to the action of two forces \vec F_1 и \vec F_2, so it is called their resultant force. When attached to the body at the same time n forces, their resultant force is equal to the vector sum of all forces  \vec F_1, \vec F_2,…, \vec F_n, applied to the body:

\vec F =\vec F_1 + \vec F_2 ++ \vec F_n

 (The information has been taken from e-book «Fundamentals of classical physics” (author V.D. Shvets), which was published in 2007 by “1C-Multimedia Ukraine” publishing company. Copyright is reserved by certificate Number 55955 from 06.08.2014. More information can be found at https://ipood-kiev.academia.edu/ValentynaShvets/Books)

TASKS

Task 1. The body slides on the inclined plane that makes an angle with the horizon 45^0 . The dependence of the traversed path occasionally looks as s(t)=Ct^2, where C=1.73 \frac{m}{s^2}. Find coefficient friction of the body and plane.

Task 2. In the horizontal plane from the rest moves the body mass 5 kg under the force 30 N, applied to the body at an angle  30^0 to the horizontal plane. The coefficient of friction between the body and the plane is equal to 0.2.   Determine the path traversed by the body for 10 seconds.

Task 3. The plane describes a circle of radius 800 m at a speed of 720 \frac {m}{s}. Find the amount of force with which the pilot presses on the seat at the top of the circle.

Task 4. The body slips on inclined-plane length L and the angle \alpha with constant speed. Find time during this body slips on this plane, if the angle is equal to \beta, where \beta > \alpha.

Task 5. On the block of 20 cm radius, which has a moment of inertia of 50 kgm^2 is thin  thread. To the ends of the thread suspended from two bodies, whereby a block is rotated with an angular acceleration of 2.36 \frac {rad}{s^2}. Find the force causing the rotation unit when the friction torque between the thread and the block is equal to 98.1 Nm.

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