A table is supported against a wall forming an angle with the vertical. The coefficient of static friction between the table and the floor is (μe). The frictional force of the wall on the board may be neglected.
Find an expression for the maximum value that the angle can take without the table slipping on the floor.
Well, isn't this a slippery situation? Let's see if I can slide in with an answer for you.
To find the maximum angle without the table slipping on the floor, we need to consider the forces acting on the table. The weight of the table can be split into two components: the component perpendicular to the floor and the component parallel to the floor.
The perpendicular component of the weight is responsible for holding the table against the wall. Let's call this force W⊥. The parallel component of the weight tries to make the table slip on the floor. We'll label this force W∥.
The static frictional force is what prevents the slipping, so we can equate the maximum static friction force to the parallel component of the weight:
μe(W⊥) = W∥
W⊥/W∥ = 1/μe
Now, we can use some trigonometry to relate the components of weight to the angle θ between the table and the vertical. The perpendicular component of the weight can be expressed as:
W⊥ = W * cos(θ)
And the parallel component of the weight is:
W∥ = W * sin(θ)
Plugging these into our expression for the ratio of the components, we get:
(cos(θ))/(sin(θ)) = 1/μe
Simplifying, we can rewrite this as:
cot(θ) = 1/μe
Taking the arctangent of both sides, we find:
θ = arctan(1/μe)
So, the maximum value that the angle θ can take without the table slipping on the floor is given by:
θ = arctan(1/μe)
Remember, this expression assumes that the frictional force between the wall and the table is negligible. Now, I hope I didn't slip up while explaining, and that this answer brings a smile to your face!
To find the maximum value of the angle at which the table does not slip on the floor, we can use the concept of equilibrium. When the table is at the maximum angle without slipping, the frictional force between the table and the floor will be at its maximum value.
Let's denote the maximum angle as θ (theta).
In order to prevent the table from slipping, the component of the weight of the table perpendicular to the floor (mgcosθ) must be balanced by the normal force(N) from the floor.
N = mgcosθ
The frictional force between the table and the floor is given by the product of the coefficient of static friction (μe) and the normal force:
Frictional force (F) = μeN
Since the frictional force opposes the component of the weight parallel to the floor (mgsinθ), we can write:
F = mgsinθ
Setting the maximum frictional force equal to the maximum component of the weight parallel to the floor, we have:
μeN = mgsinθ
Substituting N = mgcosθ, we get:
μe(mgcosθ) = mgsinθ
Dividing both sides by mg, we have:
μe cosθ = sinθ
Now, we can rearrange the equation to solve for θ:
μe cosθ = sinθ
Divide both sides by cosθ:
μe = tanθ
Finally, calculate the inverse tangent of both sides to find θ:
θ = arctan(μe)
So, the maximum angle θ at which the table does not slip on the floor is given by the arctan of the coefficient of static friction (μe).
To find the maximum value that the angle can take without the table slipping on the floor, we can set up a force balance on the table.
Let's denote:
- θ: angle between the table and the vertical
- Fg: gravitational force acting on the table (force due to its weight)
- Fn: normal force exerted by the floor on the table
- Ff: static frictional force between the table and the floor
The vertical component of the gravitational force is given by Fg * sin(θ), and the horizontal component is given by Fg * cos(θ).
Since the table is not slipping on the floor, we know that the static frictional force is equal to the horizontal component of the gravitational force, i.e., Ff = Fg * cos(θ).
The maximum value that the static frictional force can reach is given by the equation Ff = μe * Fn, where μe is the coefficient of static friction. Therefore, we can write:
Fg * cos(θ) = μe * Fn
Next, we need to express the normal force (Fn) in terms of the gravitational force (Fg). The normal force is equal to the vertical component of the gravitational force, i.e., Fn = Fg * cos(θ).
Substituting this into the previous equation, we have:
Fg * cos(θ) = μe * (Fg * cos(θ))
Now, let's cancel out the Fg terms:
cos(θ) = μe * cos(θ)
Dividing both sides by cos(θ), we get:
1 = μe
Therefore, the maximum value that the angle (θ) can take without the table slipping on the floor is given by θ = arccos(μe).