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    The top end of a two-by-four piece of lumber that is 4.2 m long is leaned at a height of 2.4 m against a smooth wall so that the bottom end makes an angle of 35 ∘ with the floor. This board has an inertia of 4.2 kg .

    What is the normal force exerted by the wall on the board?

    Attempted solution:

    The sum of the forces in both the x and y directions is zero. The gravitational force and the normal force from the floor are both exclusively in the y direction and the normal force from the wall and frictional force from the ground are in the x direction.

    The friction force and the normal force from the wall on the board are in opposite directions, so we know
    Fx=Fw-Ff=0
    Ff=Fw and that they are acting in opposition to each other.

    We also know that net torque is zero, since the system is not moving.
    In the x direction
    tao=Fxrsin(theta)

    Since I do not have a coefficient of static friction for the floor or for the lumber, how am I supposed to solve this question?
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    Although I do not agree with your initial assumptions, you can always enter a new variable for the floor friction or create an assumption to provide a solution.
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    I haven't attempted it, but since I don't know the friction on the floor, I'd take moments about that point - meaning any force acting there is out of the equation.

    Edited to add: and unless there's anything missing from your question (or unless I've misunderstood) then it becomes a very straightforward problem.
 
 
 
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