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Forces/Newtons Laws HW Problem

  1. Feb 22, 2005 #1
    A jet aircraft is climbing at an angle of 43° above the horizontal and is also accelerating at 4.9 m/s2. What is the total force that the cockpit seat exerts on the 65 kg pilot?(Magnitude & Direction ° above the direction of motion of the plane)

    This is what I have so far:
    Horizontal Component of this Force = 318 .5 Cos 43° = 232.9 N
    Vertical Component of this Force = 318.5 Sin 43° = 217.2 N

    Direction: arctan(217.2/232.9)=43°

    But it said all my answers are wrong, just wondering what I am doing wrong???? :confused:
  2. jcsd
  3. Feb 22, 2005 #2


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    Remember that if the airplane were not accelerating at all, the pilot would still have weight- he is exerting his weight on the seat and the seat is, of course, applying the same force to him.
  4. Feb 22, 2005 #3
    You should not look for the net force, but for the contact (normal) force that the seat applies on the pilot. The two forces therefore are [tex]\vec{N}[/tex], and [tex]\vec{G}[/tex] which is vertical. Break them down into components and solve for each component of [tex]\vec{N}[/tex].
    Last edited: Feb 22, 2005
  5. Feb 22, 2005 #4
    You didnt get the force on the passenger correct

    Keep in mind that what you've computed is the force on any object but you must keep in mind that the person on the plane has a mass and has a normal force perpenduclar to the surface of hte plane. You have to take thoise into account as well

    P.S. this has been edited because i had answered previousl being distrtacted for a long time in the between thus i sounded contradictory :tongue:
    Last edited: Feb 22, 2005
  6. Feb 22, 2005 #5
    Are you contradicting yourself: saying that the force on the passenger is incorrect, but the component forces are correct???
    The problem asks what it asks: the horizontal and vertical components of the normal force the seat applies on the pilot.
  7. Feb 26, 2005 #6
    I'm also working on this problem...My given numbers are a=4.7m/s^2 angle=49degrees mass=73kg.

    Wouldn't the total force that the seat exerts on the pilot be defined by finding the resultant of the following 2 acceleration vectors?:

    Vertical component acceleration + gravitational acceleration:
    [4.7*sin(49) + (9.8)]


    Horizontal component of acceleration

    I get a resultant magnitude of 13.65m/s^2 (which equals 996N) at 77degrees above the horizontal (hmm :confused: doesn't seem right at all.) Please Help.
    Last edited: Feb 26, 2005
  8. Feb 28, 2005 #7
    Why, doesn't your answer look OK? An angle of 77 degrees above the horizontal for [tex]\vec{N}[/tex] seems all right, because when combined with the vertically downward [tex]\vec{G}[/tex] it may well result in a net force whose angle with the horizontal is 42 degrees.

    Your reasoning with accelerations is not good, because [tex]\vec{a}[/tex] is the net acceleration and you cannot add g, the free fall acceleration. After all, it is not the net force that is asked, but the force that the seat applies on the pilot!

    So, just reason with forces, breaking them into x and y components:

    [tex]N_x = ma_x = macos\theta[/tex], and

    [tex]N_y - G = ma_y = masin\theta[/tex]
    [tex]N_y = masin\theta + mg[/tex]
    [tex]N_y = m(asin\theta + g)[/tex]

    then, apply the Pythagorean theorem to find the magnitude of N and arctan to find the angle [tex]\phi[/tex] that N makes with the horizontal.
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