Question

As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

Answer 1

In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.

The horizontal component of this force is given as

F_x = Fcos(6.7)

While the vertical component of this force would be

F_y = Fsin(6.7)

In the vertical component, the sum of Force indicates that:

$\sum F_y= 0$

The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:

$N = mg+Fsin(6.7)$

In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:

$\sum F_x = 0$

$F_x = F_{friction}$

$Fcos (6.7) = N\mu$

Using the previously found expression of the Normal Force and replacing it we have to,

$Fcos(6.7)= \mu (mg+Fsin(6.7))$

Replacing,

$Fcos(6.7)= (0.87) (mg+Fsin(6.7))$

$Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))$

$Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)$

$F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)$

$F = \frac{0.87 (mg)}{(cos(6.7)-0.87sin(6.7))}$

$F = \frac{0.87(128000*9.8)}{(cos(6.7)-0.87sin(6.7))}$

$F = 1.95*10^6N$

Finally the acceleration would be by Newton's second law:

$F = ma$

$a = \frac{F}{m}$

$a = \frac{ 1.95*10^6}{128000}$

$a = 15.234m/s^2$

Therefore the greatest acceleration the man can give the airplane is $15.234m/s^2$