Question

Two vertical walls are separated by a distance of 1.5 m. Wall 1 is smooth, while wall 2 is not smooth. A uniform board is propped between them. The coefficient of static friction between the board and wall 2 is 0.98. What is the length of the longest board that can be propped between the walls?

Answer 1

The length of the longest board that can be propped between the walls can be determined by considering the balance of forces acting on the board. Using the given coefficient of static friction and the weight of the board, the maximum length is calculated to be approximately 0.1547 meters.

Step-by-step explanation:

The length of the longest board that can be propped between the walls can be calculated by considering the balance of forces acting on the board. The maximum length occurs when the board is on the verge of slipping. In this case, the static friction force between wall 2 and the board is at its maximum value and is equal to the maximum static frictional force that can be exerted, given by the equation:

Frictional force = coefficient of static friction * Normal force

Since the board has uniform weight distribution, the normal force acting on wall 2 is half of the weight of the board. Thus, setting the frictional force equal to the maximum static frictional force:

Coefficient of static friction * (Weight of board / 2) = Maximum static frictional force

Solving for the weight of the board:

Weight of board = 2 * (Maximum static frictional force / Coefficient of static friction)

Substituting the given values:

Weight of board = 2 * (0.98 * 9.8)

Weight of board = 19.24 N

Since the weight of the board is known, the length of the longest board can be determined using the equation:

Length of longest board = Distance between walls + 2 * Overhang

Where the overhang is the portion of the board on either side of wall 2. Since the board is symmetrically placed between the walls, we can assume equal overhangs. Therefore:

Overhang = (Length of longest board - Distance between walls) / 2

Substituting the values:

Overhang = (L - 1.5) / 2

Plugging in the calculated weight of the board:

Overhang = (19.24 * L / 2 - 1.5) / 2

Using the constraint that the board is on the verge of slipping, the maximum static frictional force can be equal to the weight of the board:

Maximum static frictional force = Weight of board

From above, we already calculated the weight of the board to be 19.24 N. Setting this equal to the maximum static frictional force:

0.98 * Normal force = 19.24

Thus, the normal force acting on wall 2 is:

Normal force = 19.24 / 0.98 = 19.67 N

Substituting this in the formula for overhang:

Overhang = (19.24 * L / 2 - 1.5) / 2 = 0

Simplifying the equation:

19.24 * L / 4 - 1.5 / 2 = 0

19.24 * L / 4 = 1.5 / 2

19.24 * L / 4 = 0.75

L = (4 * 0.75) / 19.24 = 0.1547

Therefore, the length of the longest board that can be propped between the walls is approximately 0.1547 meters.

Answer 2

The length of the longest board that can be propped between the walls is approximately 2.23 meters.

Step-by-step explanation:

To find the length of the longest board that can be propped between the walls, we need to determine the maximum angle at which the board will not slip. The coefficient of static friction between the board and wall 2 is given as 0.98. The angle θ at which the board will not slip is equal to the arctan of the coefficient of static friction.

θ = arctan(μ)
θ = arctan(0.98)
θ ≈ 44.14°

To find the length of the board, we can use the formula:
length = distance / sin(θ)
length = 1.5 / sin(44.14°)
length ≈ 2.23 m