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Block against a wall

  1. Oct 10, 2013 #1
    A 1.0 kg wood block is pressed against a vertical wood wall by a 12 N force (into the wall, at a 30 degrees angle up from the horizontal). if the block is initially at rest, will it move upward, move downward, or stay at rest?

    I'm working on the problem above, and through Google found this link. I understand how to accomplish the problem. However, I'm having trouble identifying the forces in play. My teacher wants us to accomplish these sort of problems a very specific sort of way.
    1) Identify forces (In this case I'd have horizontal forces and vertical forces)
    2)Draw a free-body diagram
    3) Apply Newton's second law in Component form. (ƩFx=max/ƩFy=may)

    And solve.

    So far my horizontal force is the Push force (30cos12) and my vertical forces are Static Friction, weight and also Push force (30sin12).

    "Fnet(x) = Fn - Fpush(x) , and since Fnet(x) is 0, then Fn = Fpush (x)" (from link)
    I'm not understanding where he is getting Fn from or what Fn even stands for, I thought perhaps it was normal force, but there is no horizontal surface the block of wood is on, so there should be no normal force.
     
  2. jcsd
  3. Oct 10, 2013 #2
    There isn't just one horizontal force acting on the block. There is another force. Fn is the horizontal force exerted by the wall on the block. Similarly, the static friction force is the vertical force exerted by the wall on the block.
     
  4. Oct 10, 2013 #3
    Yea, I knew there wasn't just the one horizontal force, that's just what I had come up with so far. I thought that Fn was normal force, which only applied to horizontal surfaces.

    So normal force is any force that opposes the force on an object by another object in the opposite direction regardless of being horizontal or vertical? Or is not normal force at all, and is something that eludes me?
     
  5. Oct 10, 2013 #4
    Normal force is defined as the component of the contact force perpendicular to the interface.
     
  6. Oct 12, 2013 #5
    So basically... "normal force is any force that opposes the force on an object by another object in the opposite direction regardless of being horizontal or vertical" in layman terms?

    Even then, how do you have a normal force in a horizontal plane when there is no object pressing against it in the horizontal plane? Unless resolving the force applied to the block pressed against the wall (12N) in the x and y axis gives you the normal force in each plane?
     
  7. Oct 12, 2013 #6
    The block is pressed against the wall with a horizontal component of force Fcos30° and pushed upwards with a component Fsin30°.The block in turn pushes against the wall .There is contact force between the block and the wall .The component of the contact force perpendicular to the wall (looks horizontal to us) is the normal force .the component of contact force parallel to the wall (looks vertical to us) is known as friction which opposes relative motion between the block and the wall.
     
  8. Oct 12, 2013 #7
    Let's be clear. There are three forces acting on the block:
    1. Weight of the block
    2. 12 lb. force applied to the block from the right
    3. Force exerted by the wall on the block from the left

    We know that the wall is exerting a force on the block because, otherwise, the block would go through the wall.

    The force exerted by the wall on the block can be resolved into two components: one component perpendicular to the block/wall interface, and the other component tangent to the block/wall interface. The component perpendicular to the interface is called the Normal component of the Force, or, simply the Normal Force. The component tangent to the interface is called the Shear Force or Frictional Force.

    Chet
     
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