🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning



Numerade Educator



Problem 80 Hard Difficulty

A man seeking to set a world record wants to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to the airplane. The mass of the man is 85 kg, and the coefficient of static friction between his shoes and the runway is 0.77. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.


$5.9 \times 10^{-3} \mathrm{m} / \mathrm{s}^{2}$


You must be signed in to discuss.

Video Transcript

to solve this question are you choose the following reference frame the vertical axis, which is the why access we'll be pointing up on the horizontal access, which I call the X axis will be pointing to the right. Then we use Newton's second law to calculate what is the acceleration off the airplane. I noticed that the airplane moves on the horizontal axis, so we should applying Newton's second law on that access to discover its acceleration. By doing that, we get the following. So applying Newton's second law to the plane specifically to the X axis, we got that the net force The X direction is He goes to the mass off the airplane times acceleration off the airplane in that direction. The net force in the direction is composed only by one force. The tension force. Therefore, the tension force. Is it close to the mask off the airplane times its exploration? Then the acceleration off the airplane is It goes to the tension divided by its mess. Why have I done that? Because of the following. We want to know what is the maximum possible acceleration for the airplane. Then we know now that acceleration is proportional to the tension. Notice that we can't really change the mass off the airplane, so the only thing we can change is the tension force. So now our objective is to find what is the maximum tension force because the maximum tension force we will result in the maximum acceleration. So now how can you get her mind the maximum tension force? To do that, we have to look now at the man that is trying to pull the airplane. Specifically, we should apply Newton's second law to the man. But now, as the airplanes moving on the horizontal axis, that is, apply Newton's second law on the men on the horizontal axis first. So now we are applying Newton's second law to the men on the X axis. Its results in the following the Net force and acts access. Is it close to the mass off the man times? He's acceleration next direction, but the net force and the extra actions composed by true forces, the tension force and the Frictional force notice that the tension force that is acting on the man points the left. Then it has a minus sign, while the frictional force point to the right so it has a positive sign than the frictional force. Minus attention force is the mass off the man times. He's acceleration in the extraction. Now note. The following acceleration off the men is the same as acceleration off the airplane. This is because we are supposing that the cable that has been use it is an ideal cable. It means that it can't be stretch it or contracted. Then we can equate acceleration off the airplane with the acceleration off the men to get the following the Frictional Force Minors Addiction Force is the mass off the man times his acceleration which is the tension force divided by the mass off the airplane. Then we can send the station force to the other side. To get fictional force is equals to attention force plus the mass The man divided by the mass off the plane time Attention force. We can then factor detention force on the right inside to get attention Force times one plus mass off the man divided by miles off the airplane. Is it close to the frictional force? Then we can so for detention. For us to get that detention force is the frictional force divided by one plus the mass off the men divided by the mass off the airplane. And now we can conclude the following The maximum tension happens when the frictional force is also the maximum fictional force. And you know that the maximum frictional force this is a top where the frictional force is static is given by the static frictional coefficient times the normal force that acts on the men because the frictional force is acting on the men. Then it was they divided by one plus the mass off the man divided by the mass off the airplane. Let me organize the board to continue serving this question. Now the only thing that is left for us to discover is the magnitude off the normal force that acts on the men. But this is very easy to do because you see that the normal force acts on the vertical axis and the normal force that acts on the men is being counterbalanced by the weight force that acts on the men. So as the man isn't going to fly away nor going to fall below the floor, the only conclusion that it's possible is that the normal force has the same magnitude as the weight force. Then lactation force can be reading us frictional coefficient times the weight force that acts on the men, which is the mass off the man times that celebration of gravity divided by one plus the mass off the men divided by the mass off the airplanes. Now we played in the values that were given in the problem, remembering that the acceleration of gravity is approximately 9.8 meters per second squared. Then the maximum possible tension force is given by 0.77 times 85 times 9.8 divided by one plus 85 divided by 109 3 zeros. And these gives us attention force off approximately 640.9 beauticians. Now we can go back to the equation for the acceleration of the plane, which is this one calculates the maximum possible acceleration. So the acceleration and the exact seals off the plane is the question is 640.9, divided by the mass off the airplane, and these results in an acceleration off approximately zero 0.0 59 meters per second squared, which we can write as 5.9 times 10 to the minus tree meters per second squared

Brazilian Center for Research in Physics
Top Physics 101 Mechanics Educators
Christina K.

Rutgers, The State University of New Jersey

Marshall S.

University of Washington

Zachary M.

Hope College

Aspen F.

University of Sheffield