1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please Help Me With Force Problems

  1. Sep 22, 2005 #1
    I'm having real trouble working through force problems. I can set up the free body diagrams fine and I understand the situations, but our instructor does not do well with explaining the math parts of problems. I have a few force problems that I just cannot work through. Can anyone help me?

    One is: a) Two boats pull a 75 kg water skier. If each boat pulls with 600 N and the skier travels at a constant V, what is the magnitude of the retarding force between the water and the skis? b) if each boat pulls with 700N, what is the magnitude of the skier's acceleration?

    I am thinking that I need to resolve this into y and x components but I really have no idea?

    Another is: a) a 65 kg water skier is pulled by a boat with a horizontal force 400 N due east with water drag of 300N. A gust of wind supplies another horizontal force of 50 N at an angle of 60 degrees north of east. What is the skierr's acceleration at that instant? b) what would be the skiers acceleration if the wind force were in the opposite direction to that in part a?

    I am just as lost as on the first one for this as they are similar problems.

    The last one is: a .2 kg ball is released from a height of 10m above the beach and it leaves an impression 5 cm deep, what is the average force acting on the ball by the sand?

    I jsut have no idea where to go with any of these. I don't even care if I have the answers here, I would just as soon take some advice and tutorials on how to do force problems as the book provides minimal help, as does my instructor. Any help on any of these would help as I am completely frustrated. :mad:
  2. jcsd
  3. Sep 22, 2005 #2
    First of all, I think that's impressive. I've never seen anyone ski being pulled by two boats. Wow. Assuming that the boats are directly in front of the skier, and not pulling at an angle to the skiers direction, I think you can treat this as one big boat pulling one big ski at 1200 N.

    It would have been good if you could have given some idea of how you drew your free body diagrams. But I'll try my hand at helping you with (a), as best as I can.

    When you say you need to separate this into X and Y components, you are right. If you have drawn two free body diagrams, then you have already done this, and you just need to translate it into mathematics.

    For the forces in the Y direction, you have a force of unknown magnitude pointing up (call it, N for Normal), and the weight of the skier down (that's mg). Call up positive, down is negative. And the skier isn't accelerating in the Y direction, he just stays on the water, since he is an ideal skier. That means that when you add these forces together, they have to equal zero (since F=ma). So [tex] \sum F_y = N - mg = 0 [/tex] If he had been accelerating, you would have written [tex] \sum F_y = N - mg = ma [/tex].

    The situation is the same for the X forces. You have the force of the boat moving the skier in the positive X direction, and a resistive force pushing back on the skis, that's aiming in the negative X direction. The skier is moving, but he is not accelerating, so these forces sum to zero also. Call the resistive force [tex]f_r[/tex] and the force of the boats [tex] f_b [/tex]. I'll leave it to you to write the equation.

    You can probably just treat the water as you would resistance. That's proportional to the normal force, so you have a third equation [tex] f_r = \mu N [/tex]. This gives you three equations in three unknowns, and you should be able to solve this pretty easily, and if neither of us made any mistakes, then you should have your solution.

  4. Sep 22, 2005 #3
    Ah yes see that is what I was trying. I'm sorry I didn't put my ideas about my free body diagram, but I was kind of in a panic.

    Anyway, the thing about the problem is that the two boats are at an angle of 45 degrees from the skier. I really can't believe I forgot to include that. I don't know how to integrate that into the problem.
  5. Sep 22, 2005 #4


    User Avatar
    Gold Member

    The fact that the ropes are at 45° to the force means that the two forces don't just add in magnitude; you have to add thier components. If the boats pull the skier in the +y direction, then you can see that the x-components of the forces cancel with each other. The force from each boat in the y-direction is 600cos(45°) from trigonometry.
  6. Sep 22, 2005 #5
    for a) ud have to resolve horizontally ( resolving vertically will result in an over all zero force acting) since its at constant velocity the force dragging the ski forwrd will equal the retardation force
  7. Sep 22, 2005 #6
    o bother u got there before me lol
  8. Sep 22, 2005 #7
    So you would take the 600cos45 and that would be the force or would that be multiplied by two? There is one boat above the +x axis pulling at a 45 degree angle from the skier and one below the +x axis pulling the skier so this is making the problem confusing. I am thinking then it would just be 660cos45?

    for part b would you just take the 700cos45=75a and solve for a that way or is there more to it?

    Can anyone help me with the other two? I just feel like I am in a fog... :confused:

    For part a on the second questin i said 50cos68=18.7N 50sin68=46.4N w=mg=65kg(9.8m/s^2)=637 N... is this in any way in the right direction?
    Last edited: Sep 22, 2005
  9. Sep 22, 2005 #8
    Can anyone point me in any direction whatsoever?
  10. Sep 22, 2005 #9

    Doc Al

    User Avatar

    Staff: Mentor

    For part a the idea is that since the skier is moving at constant speed, the net force must be zero. The forces acting on the skier are (1) the force of the ropes, pulling forward and (2) the retarding force of the water, pulling backward. The forward force of each rope is [itex]600 \cos(45)[/itex].

    Once you find the retarding force in part a, you use it to solve part b. In part b the skier is accelerating. Once again, the forces are (1) the force of the ropes, pulling forward and (2) the retarding force of the water, pulling backward. Assume that the retarding force is the same, but the ropes are pulling harder. So find the net force and use F=ma to find the acceleration.
  11. Sep 22, 2005 #10
    You should have two vector arrows pointing to the right, parallel to the X axis, that represent the forward component of motion for the skier. Opposite these, you should have two vector arrows that point to the left, that represent the resistance of the water (because that opposes the motion of the skier). So, call each of the forward arrows [itex] F_x [/itex] and each of the backward arrows [itex] F_r [/itex]. If an arrow points in a positive direction, add it, if it points in a negative direction, subtract it:

    [tex] F_x + F_x - F_r - F_r = 0 [/tex]

    You also have an arrow along (or parallel) to the Y axis, pointing down, and one pointing up. These represent the forces that are keeping the skier in the middle of the two boats. Call these [itex] F_y [/itex]. Again, positive arrows you add, and negative arrows you subtract:

    [tex] F_y - F_y = 0 [/tex]

    So these forces cancel out, and you can forget about them. Returning to our first equation:

    [tex] 2F_x - 2F_r = 0[/tex]

    [tex] 2F_x = 2F_r [/tex]

    This tells you that the force moving the skier forward is exactly countered by a force moving the skier backwards. All that remains is to use trigonometry to get the numeric value [itex] F_r [/itex] and decide if you need the total force (net force) of resistance, or just the resistance for one boat.

    The rest of the problems can be solved in this same way, just proceeding methodically, step by step, without panic.

  12. Sep 22, 2005 #11
    My only problem with the last one is that I don't understand what forces are working. There are probably two downward forces due to the ball's downward force and gravity and Normal force upward but there are no values for any of these so I am not sure how to proceed.
  13. Sep 22, 2005 #12

    Doc Al

    User Avatar

    Staff: Mentor

    There are only two forces acting: (1) The downward force of gravity (the ball's weight), and (2) the upward force of the ground on the ball. If you need the ball's weight, you can calculate it from its mass. The upward force is what you are asked to find.
  14. Sep 22, 2005 #13
    So for the first one I took Fcos45 = 600 cos45 Fr=424.3N (must be negative)
    then part b) 700cos45=75a, a=6.6m/s^2... is this the correct way? The way I'm reading your solutions it seems as if that's what you're saying? It makes sense to me that way at least.

    The next one I took 50cos60=25N 100+50cos60=125 125/65=2 m/s^2 which was the answer in the book... is that in any way correct?

    For the last one I seem to understand better now. I took .2kg(9.8)=1.96, 10m/.05m=200, 200x1.96=392 which is the answer in the book... I wonder if that is the correct method because I udnerstand it but I'm not sure about the math. Our instructor taught us the concepts but very little math. That is why I guess I am thinking too hard on the concepts and am struggling with what is pretty much simple algebra... Are these methods correct?
  15. Sep 23, 2005 #14

    Doc Al

    User Avatar

    Staff: Mentor

    There are two boats pulling; 600 cos45 is just the force from one of them. (Double it.)
    No. Find the net force on the skier: Rope force - water retarding force. Set that net force equal to ma.

    It looks OK to me: Before the gust of wind, the net horizontal force was 400 - 300 = 100N east. The wind adds another 50cos60 = 25N going east, so the net force is now 125N east.

    Again, it's hard to say if your thinking is correct, since all you show are a few equations. (Since you don't explain what principles you are applying, you could have just gotten lucky for all I know. :smile: ) Nonetheless, setting the KE of the ball at impact (KE = mgh, from conservation of energy) equal to the work done to bring the ball to rest by the force of the ground (W = Nd), gives mgh = Nd, or N = mgh/d. (This ignores the small bit of PE due to the ball lowering an additional 5cm; don't worry about it.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Please Help Me With Force Problems
  1. Force help me please =) (Replies: 11)