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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PatchworkTeapot
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- 20-02-2018 03:55
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Apocalyps
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- 20-02-2018 04:13
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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- 20-02-2018 16:04
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.Last edited by phys981; 20-02-2018 at 16:10.
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