Solution D1-55 (Figure D1.5 condition 5 S.M. Targ 1989)

Solution to problem D1-55 (Figure D1.5, condition 5, S.M. Targ, 1989)

There is a load of mass D that moves in a curved pipe ABC located in a vertical plane. Pipe sections can be inclined or horizontal (see Figures D1.0 - D1.9 and Table D1). At point A, the load receives an initial speed v0. In section AB, in addition to the force of gravity, the load is acted upon by a constant force Q (its direction is indicated in the figures) and a resistance force of the medium R, which depends on the speed v of the load and is directed against the movement. The friction of the load on the pipe in section AB is not taken into account. At point B, the load moves to section BC of the pipe without changing its speed, where it is acted upon by a friction force (coefficient of friction of the load on the pipe f = 0.2) and a variable force F, the projection of which Fx on the x axis is given in the table. The load is considered a material point. We denote the distance AB by l, the time of movement of the load from point A to point B by t1. We will neglect the friction of the load on the pipe in the section BC.

It is necessary to find the law of cargo movement on the aircraft section, i.e. expression for the x coordinate of point D depending on time t, i.e. x = f(t), where x = BD.

The solution to this problem can be divided into two parts: the movement of the load in section AB and the movement in section BC.

  1. Movement of cargo on section AB.

The load is acted upon by the force of gravity D, the constant force Q and the resistance force of the medium R. Newton’s second law for the load has the form:

D - Q - R = Dv,

where v is the speed of the load.

Considering that the resistance force of the medium R is proportional to the speed v, i.e. R = kv, where k is some constant, we get:

D - Q - kv = Dv.

Thus, the equation for the movement of cargo in section AB has the form:

Dv + kv = D - Q.

Let us introduce the following notation:

a = k/D, b = Q/D.

Then the equation of motion can be written as:

v' + av = 1 - b,

where v' = dv/dt.

The solution to this linear differential equation is:

v = (1 - b)/a + Ce^(-at),

where C is the integration constant, which can be found from the initial conditions. At point A, the load has an initial speed v0, therefore, C = (v0 - (1 - b)/a).

Thus, the speed of the load in section AB is equal to:

v = (v0 - (1 - b)/a)e^(-at) + (1 - b)/a.

By integrating the speed, we find the law of movement of the load in section AB:

x = l - (1 - b)/a - ((v0 - (1 - b)/a)/a)(1 - e^(-at)) - (1 - b)/a*t - (1/2)k/at^2.

  1. Movement of cargo on the aircraft section.

The load is acted upon by gravity D, variable force F and frictional force. Newton's second law for load has the form:

D - F - fN = Dv',

where N is the normal force, f is the friction coefficient.

Considering that the normal force is equal to the weight of the load on the aircraft section, i.e. N = D, and the projection of force F on the x axis, equal to Fx, can be expressed in terms of time t, we get:

F = Fx(t).

Thus, the equation for the movement of cargo on the aircraft section has the form:

Dv' = D - Fx(t) - fD.

Solving this differential equation, we find the law of movement of the load on the aircraft section:

x = l + v*t - (1/2)Fx(t)/Dt^2 + (1/2)ft^2.

So, the law of cargo movement on the aircraft section has the form:

x = l + v*t - (1/2)Fx(t)/Dt^2 + (1/2)ft^2.

Thus, the complete law of movement of cargo in section ABC can be written as:

x = l - (1 - b)/a - ((v0 - (1 - b)/a)/a)(1 - e^(-at)) - (1 - b)/at - (1/2)k/at^2 + l + vt - (1/2)Fx(t)/Dt^2 + (1/2)ft^2.

Ответ: x = 2l - (1 - b)/a - ((v0 - (1 - b)/a)/a)(1 - e^(-at)) - (1 - b)/a*t - (1/2)k/at^2 - (1/2)Fx(t)/Dt^2 + (1/2)ft^2.

The product “Solution D1-55 (Figure D1.5 condition 5 S.M. Targ 1989)” is a digital product that represents a solution to a problem from S.M.’s textbook. Targa on mechanics. The solution contains a detailed description of how to solve this problem, as well as formulas and calculations that will help you understand the solution process.

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This digital product will be useful for students, teachers and simply people interested in mechanics. By gaining access to this product, you can improve your knowledge of mechanics and learn how to solve similar problems.


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Solution D1-55 describes the movement of a load of mass m, which receives an initial speed v0 at point A and moves along a curved pipe ABC located in a vertical plane. In section AB, in addition to the force of gravity, the load is acted upon by a constant force Q and a resistance force of the medium R, which depends on the speed of the load. At point B, the load moves to section BC, where, in addition to the force of gravity, it is acted upon by the friction force and the variable force F. The coefficient of friction of the load on the pipe is f = 0.2.

To find the law of movement of the load in section BC, it is necessary to know the distance AB=l or the time t1 of movement of the load from point A to point B. The friction of the load on the pipe in section AB is neglected.

The task is to find the function x=f(t), where x=BD is the distance from point B to point D, and t is the time of movement of the cargo on the aircraft section. To solve the problem it is necessary to use data from the table and figures D1.0-D1.9.


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