# 3 Problems related to Vectos: Force, Motion, currents

Please explain in step by step detail the following:

25. A weight of 850 pounds is suspended by two cables. One cable makes an angle of 66 degrees with a vertical line, the other makes an angle of 42 degrees with a vertical line. Find the amount of force exerted by each of the cables.

24. An airplane is scheduled to reach an airport 600 miles away in exactly 3 hours. If the airport is directly northeast of the starting point and a wind of 20 mph will be forcing the plane southward, at what angle should the pilot head the plane? What should the air speed be?

23. The resultant of two currents of electricity is 12 amperes at a direction angle of 30 degrees. If one of the currents consists of 15 amperes at 60 degrees ( from what would be the positive x axis), what is the vector of the other current?

On the above please explain resultant, vector, and formulas used.

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69661

25. A weight of 850 pounds is suspended by two cables. ONe cable makes an angle of 66 degrees with a vertical line, the other makes an angle of 42 degrees with a vertical line. Find the amount of force exerted by each of the cables.

Please give all formulas used.

Answer:

The force exerted by the cable is due to the tension in the cables.

Let the tensions in the cables or the force exerted by the cables are F1 and F2 in magnitude, the directions are given. The two forces balance the weight of the body suspended, means that the resultant of F1, F2 and W must be zero.

Force is a vector quantity and to find the resultant of number of vectors, we write them in components form and then add the components in each direction algebraically. Here i and j are unit vectors in x and y directions respectively.

If vector A makes an angle Î¸ with positive x axis then the components Ax and Ay are given by A cos Î¸ and A sin Î¸ respectively. Hence the three vectors can be written as

and

As the body is in equilibrium under the three forces, their resultant should be zero and hence the corresponding components must be zero, gives

Solving these equations we can calculate the magnitudes of the two forces as

From equation 1

F1 = (0.545/0.913) F2

Or F1 = 0.597 F2 ............3.

Substituting the values in equation 2 we get

0.407*0.597 F2 + 0.839 F2 = 850 pounds

or 1.136 F2 = 850

or F2 = 748.3 pounds.

Substituting this in equation 3 we get

F1 = 0.597*748.3 =446.7 Pounds.

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24. An airplane is scheduled to reach an airport 600 miles away in exactly 3 hours. If the airport is directly northeast of the starting point and a wind of 20 mph will be forcing the plane southward, at what angle should the pilot head the plane? What should the air speed be?

Answer:

The airplane is having two velocities (vectors) simultaneously, the velocity v1 relative to the air provided by the engine of the airplane (air speed) and the other va is the velocity with which wind drags it. The resultant of the two velocities is the actual velocity of the plane relative to the earth.

Let the velocity of the engine set by the pilot is

The wind velocity is

and the resulting velocity required is given by

Now as we know

gives

or vx = 141.42

and vy -20 = 141.42 or vy = 161.42

Hence the air speed of the plane should be

mph,

and in the direction Î¸ such that tanÎ¸ = vy/vx =161.42/141.42 = 1.1414

or Î¸ = 48.780 north of east.

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23. The resultant of two currents of electricity is 12 amperes at a direction angle of 30 degrees. If one of the currents consists of 15 amperes at 60 degrees ( from what would be the positive x axis), what is the vector of the other current?

Answer:

(Though the electric current is not a vector quantity, I think this problem is to learn the vectors and considering current as a vector.)

The resultant current is given in component form by

the first current is given by the equation

let the second current is

then as

we get =( )+( )

or =

Equating the corresponding vectors we get

Ix = 12cos300 - 15 cos600 = 10.39 -7.5 = 2.89 amperes

And Iy = 12sin300 - 15 sin 600 = 6.00 - 12.99 = -6.99 amperes.

Hence the magnitude of the second current is

amperes and in the direction

Î¸ = tan-1(-6.99/2.89) = tan-1(-2.4186) = -67.530

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