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More Physics help .

  1. Mar 31, 2004 #1
    More Physics help.....

    Which formulas should i use??? thankssssssssss

    1. Three forces act on a moving object. One force has a magnitude of 86.3 N and is directed due north. Another has a magnitude of 67.0 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express the direction as an angle with respect to east.

    2. The drawing shows a circus clown who weighs 772 N. The coefficient of static friction between the clown's feet and the ground is 0.545. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself? *** click on the link for pic**** http://publish.hometown.aol.co.uk/hyen84/images/untitled.bmp

    3. A person is trying to judge whether a picture (mass = 1.74 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.71. What is the minimum amount of pressing force that must be used?

    4. A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 31.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 49.9 kg, and the coefficient of kinetic friction between the skis and the snow is 0.232. Find the magnitude of the force that the tow bar exerts on the skier.
     
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  3. Mar 31, 2004 #2

    HallsofIvy

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    "1. Three forces act on a moving object. One force has a magnitude of 86.3 N and is directed due north. Another has a magnitude of 67.0 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express the direction as an angle with respect to east."
    Add the two given forces. You can do that either by analyzing the triangle made by the two and their resultant (that's pretty easy since it is a right triangle) or by setting up a coordinate system and writing them in terms of their components (also fairly easy if you take y "north" and x "east"). Since the "object" is not accelerating, the third force must the exact opposite of the sum of the two given forces.

    "2. The drawing shows a circus clown who weighs 772 N. The coefficient of static friction between the clown's feet and the ground is 0.545. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself? *** click on the link for pic**** http://publish.hometown.aol.co.uk/h...es/untitled.bmp"

    Fairly straight forward. Since the clowns weight is 772 N and the coefficient of static friction is .545, he must pull with force (772)(0.545) to overcome the static friction.

    "3. A person is trying to judge whether a picture (mass = 1.74 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.71. What is the minimum amount of pressing force that must be used?"

    This is really the same as problem 2! The whole point is that the static friction is the normal force times the coefficient of friction. In this case you want to make the static friction, 0.71*"pressing force", equal to the weight of the picture. Since the mass of the picture is 1.74 kg, its weight is 1.74(9.81)= 17 Newtons. You must have 0.71*pressing force= 17 or pressing force= 17/0.71 Newtons.

    "4. A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 31.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 49.9 kg, and the coefficient of kinetic friction between the skis and the snow is 0.232. Find the magnitude of the force that the tow bar exerts on the skier."

    The only "complication" here is the slope. The skier's mass is 49.9 kg so his/her weight is 49.9*9.81= 489 N. That's "straight down". Draw a picture and analyise the right triangle formed to see that the "normal force" on the slope is 489(cos(31))= 419.6 Newtons and so the friction force is 0.232(419.6)= 97.3 N. We also need to note that the component of weight parallel to the slope is 489(sin(31))= 251.8 N. The towbar has to equal the sum of those in order to move the skier at constant speed.
     
  4. Mar 31, 2004 #3
    Thank you sooooooooooo much..i figured it out already
     
    Last edited: Apr 2, 2004
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