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?
To find the expression for the magnitude of the normal force NW exerted by the wall on the ladder, we need to consider the forces acting on the ladder.
1) Expression for NW:
The forces acting on the ladder are:
- Weight of the ladder (mg)
- Normal force from the floor (NF)
- Normal force from the wall (NW)
- Frictional force from the floor (FF)
As the ladder is stationary, the net force in the horizontal direction is zero:
NF - FF = 0
The normal force from the floor can be expressed as:
NF = mg + FF
The frictional force from the floor can be determined using the coefficient of static friction:
FF = μ * NF = μ * (mg + FF)
Rearranging the equation, we get:
FF = (μ * mg) / (1 - μ)
We can now substitute the expression for FF into the net force equation:
NF - (μ * NF) = mg
NF - μ * (mg + FF) = mg
NF - μmg - μ * (μ * mg) / (1 - μ) = mg
(1 - μ - μ^2 / (1 - μ)) * NF = mg
(1 - μ^2 - μ) * NF = mg * (1 - μ)
NF = (mg * (1 - μ)) / (1 - μ^2 - μ)
NF = mg / (1 + μ + μ^2)
This is the expression for the normal force from the floor.
To find the expression for the magnitude of the normal force NW exerted by the wall on the ladder, we need to consider the forces in the vertical direction. The ladder is in equilibrium vertically, so the sum of all vertical forces is zero:
NW + NF - mg - M * g = 0
Simplifying the equation:
NW = mg + M * g - NF
NW = mg + M * g - (mg / (1 + μ + μ^2))
Since we know that μ = 0.41 and M is the weight of the painter (8M), we can substitute these values into the equation:
NW = (mg + 8mg) - mg / (1 + 0.41 + 0.41^2)
NW = 9mg - mg / (1 + 0.41 + 0.1681)
NW = 9mg - mg / (1.5781)
Simplifying further:
NW = (9 - 1)mg / (1.5781)
NW = 8mg / (1.5781)
NW = 5.08mg
Therefore, the expression for the magnitude of the normal force NW exerted by the wall on the ladder is NW = 5.08mg.
2) To determine the largest value of dmax for which the ladder does not slip, we need to consider the tipping point. The ladder will tip over when the net torque about the bottom end is zero.
The torque exerted by the painter's weight about the bottom end of the ladder can be expressed as:
τ = (M * g * d) * L sin(θ)
To balance the torque, we need a torque in the opposite direction, which is provided by the frictional force between the ladder and the floor.
The maximum torque provided by static friction occurs when the ladder is just at the point of slipping, which can be determined as:
τ_max = μ * NF * L
Setting the torque equation equal to the maximum torque:
(M * g * d) * L sin(θ) = μ * NF * L
Simplifying the equation:
m * d = μ * NF
Substitute the expression for NF (5.08mg):
m * d = μ * (5.08mg)
Simplifying further:
d = (μ * 5.08) meters
Substituting the value of μ (0.41):
d = (0.41 * 5.08) meters
d = 2.0848 meters
Therefore, the largest value of dmax for which the ladder does not slip is approximately 2.0848 meters.
To solve these problems, we will analyze the forces acting on the ladder.
1) The forces acting on the ladder are the weight (Mg) acting downwards, the normal force (NW) exerted by the wall on the ladder acting perpendicular to the wall, and the frictional force (Ff) between the ladder and the floor acting parallel to the floor.
Given:
Length of the ladder (L) = 5.1 m
Mass of the ladder (M) = 15 kg
Angle of the ladder with respect to the floor (θ) = 51°
Coefficient of static friction (μ) = 0.41
The normal force NW is the component of the gravitational force acting perpendicular to the wall. Using trigonometry, we can find NW.
NW = Mg cosθ
Therefore, the correct expression for the magnitude of the normal force NW exerted by the wall on the ladder is:
B) NW = 1/2Mg cotθ
2) To find the largest value of dmax for which the ladder does not slip, we need to consider the condition for equilibrium. The ladder will not slip if the frictional force (Ff) between the ladder and the floor is less than or equal to the maximum frictional force.
The maximum frictional force (Fmax) can be calculated using:
Fmax = μNW
Substituting the expression for NW from the previous question:
Fmax = μ(1/2Mg cotθ)
Now, to find the maximum distance (dmax) for which the ladder does not slip, we need to consider the torque balance around the base of the ladder. The torque due to the weight of the painter exerted at distance d from the base of the ladder must be less than the torque due to the frictional force Ff.
For equilibrium, the torques must balance:
Mgd ≤ Fmax * L
Substituting the values:
Mgd ≤ μ(1/2Mg cotθ) * L
Simplifying:
d ≤ (1/2μ cotθ) * L
Substituting the given values:
d ≤ (1/2 * 0.41 * 5.1) / cot(51°)
Calculate the right-hand side to find the maximum value of dmax:
dmax = (1/2 * 0.41 * 5.1) / cot(51°)
Therefore, the largest value of dmax for which the ladder does not slip is the calculated value of dmax.