A uniform stationary ladder of length= 5.1 m and mass= 15 kg leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. The coefficient of static friction between floor and ladder is μ = 0.41. The ladder makes an angle θ = 51° with respect to the floor. A painter of weight 8M stands on the ladder a distance d from its base.
1)Which is the expression for the magnitude of the normal force NW exerted by the wall on the ladder?
A)NW = 1/2Mg cotθ (1+16d/L)
B)NW = ½Mg cotθ
C)NW = 1/2Mg cotθ (16d/L)
D)NW = 1/2Mg tanθ (1+16d/L)
E)NW = Mg cotθ (16+d/2L)
2)What is the largest value of dmax for which the ladder does not slip, in meters?
a!!! It's A!
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1) The expression for the magnitude of the normal force NW exerted by the wall on the ladder is:
A) NW = 1/2Mg cotθ (1+16d/L)
2) To find the largest value of dmax for which the ladder does not slip, we can use the condition for static equilibrium. The ladder will not slip if the static friction force between the ladder and the floor is greater than or equal to the horizontal component of the painter's weight.
The horizontal component of the painter's weight is given by Fh = 8Mg cosθ.
The maximum friction force can be found using the equation Ffriction = μN, where μ is the coefficient of static friction and N is the normal force exerted by the floor on the ladder.
Since N = NW - Mg sinθ (because the ladder is in equilibrium in the vertical direction), the maximum friction force is given by:
Ffriction = μ(NW - Mg sinθ)
To make sure the ladder does not slip, the maximum friction force should be greater than or equal to the horizontal component of the painter's weight:
μ(NW - Mg sinθ) ≥ 8Mg cosθ
Now, substitute NW from the first expression we found for the normal force, and solve for d:
μ(1/2Mg cotθ (1+16d/L) - Mg sinθ) ≥ 8Mg cosθ
Simplify and solve for d:
1/2 cotθ (1+16d/L) - sinθ ≥ 16μ cosθ
cotθ + 16d/L - 2sinθ ≥ 32μ cosθ
16d/L ≥ 32μ cosθ - cotθ + 2sinθ
From this equation, we can isolate d:
d/L ≥ (32μ cosθ - cotθ + 2sinθ)/16
So, the largest value of dmax for which the ladder does not slip is given by:
dmax = L * [(32μ cosθ - cotθ + 2sinθ)/16]
To answer these questions, we need to analyze the forces acting on the ladder.
1) The first question asks for the expression for the magnitude of the normal force NW exerted by the wall on the ladder.
To find NW, we need to consider the forces in the vertical direction.
The forces acting vertically are:
- The weight of the ladder, which is equal to its mass multiplied by the acceleration due to gravity, Mg.
- The vertical component of the painter's weight, which we'll call Np.
Since the ladder is stationary, the sum of the vertical forces must be zero. Therefore, NW - Mg - Np = 0.
From trigonometry, we know that Np = Mgd/L, where L is the length of the ladder.
Substituting Np into the equation, we get NW - Mg - Mgd/L = 0.
Rearranging the equation, we obtain NW = 1/2 Mg (1 + 2d/L).
Now, let's look at the available options and match them with the derived expression:
A) NW = 1/2Mg cotθ (1+16d/L)
B) NW = ½Mg cotθ (incorrect, it does not include the term (1 + 2d/L))
C) NW = 1/2Mg cotθ (16d/L) (incorrect, it includes an incorrect factor of 16)
D) NW = 1/2Mg tanθ (1+16d/L) (incorrect, it uses tanθ instead of cotθ)
E) NW = Mg cotθ (16+d/2L) (incorrect, it includes an incorrect factor of 16)
Therefore, the correct answer is A) NW = 1/2Mg cotθ (1+16d/L).
2) The second question asks for the largest value of dmax for which the ladder does not slip.
For the ladder not to slip, the friction force between the ladder and the floor must be equal to or greater than the maximum possible friction force.
The maximum possible friction force is given by the coefficient of static friction (μ) multiplied by the normal force (NW). So, the maximum friction force is μNW.
The friction force can be represented as Ff = μNW, and it acts in the horizontal direction.
The force equation in the horizontal direction is: Ff - 8Mg = 0.
Substituting μNW for Ff, we get: μNW - 8Mg = 0.
From this equation, we can isolate NW and find the expression for it: NW = 8Mg / μ.
To find the largest value of dmax for which the ladder does not slip, we need to substitute the expression for NW into NW = 1/2 Mg (1 + 2d/L), and then solve for d.
Substituting NW = 8Mg / μ, we have:
8Mg / μ = 1/2 Mg (1 + 2d/L).
Cancelling out the Mg terms and rearranging the equation, we get:
16 / μ = 1 + 2d/L.
Simplifying further, we have: 2d/L = 16 / μ - 1.
Finally, solving for d, we get: d = (16 / μ - 1) * L / 2.
Now, we can substitute the given values into the expression to find the largest value of dmax.
Note: The question did not provide the value of the coefficient of static friction (μ), so it is not possible to find the exact value of dmax without that information.