Question

a uniform 4.0-m plank weighing 200.0 n rests against the corner of a wall, as shown below. there is no friction at the point where the plank meets the corner. (a) find the forces that the corner and the floor exert on the plank. (b) what is the minimum coefficient of static friction between the floor and the plank to prevent the plank from slipping?

Answer 1

The forces that the corner and the floor exert on the plank are normal forces. To calculate the minimum coefficient of static friction to prevent slipping, we set up equilibrium conditions and find it to be the tangent of the angle at which the plank rests if the center of mass is in the middle (45 degrees); in this case, the minimum coefficient of static friction would be equal to 1.

Step-by-step explanation:

To solve part (a) of the question involving a 4.0-m plank weighing 200.0 N and resting on a corner, we need to analyze the forces acting on the plank. Since there is no friction at the point where the plank meets the corner, the forces exerted by the corner (the wall and the floor) are a normal force exerted by the wall horizontally and a normal force exerted by the floor vertically. These forces create a right-angle triangle with the weight of the plank acting as the hypotenuse. The normal force exerted by the floor (F_floor) can be represented as the adjacent side of the triangle and can be found using the cosine of the angle, which in this case is a 45-degree angle since the plank is uniform and hence its center of mass is at its midpoint.

To find the minimum coefficient of static friction (part b), we can set up the equilibrium condition for the plank not slipping. This condition requires that the static friction force (f_s) needs to be greater than or equal to the horizontal component of the weight of the plank, which is the sine of the 45-degree angle multiplied by the weight of the plank.

The formula to find the minimum coefficient of static friction (μ_s) is:

μ_s = f_s / F_floor

Simplifying, we find that μ_s is equal to the tangent of the 45-degree angle (which is 1), hence μ_s = 1 if the angle is 45 degrees. If the plank center of mass is not at the middle, then the angle will be different, and we would need to calculate the tangent of that specific angle accordingly to determine μ_s.

The reaction force from the corner is the normal force, while the reaction of the floor is a combination of the normal force and the static friction force required to prevent slipping.

Answer 2

The corner and the floor exert a force of 200.0 N each on the plank in the upward direction. The minimum coefficient of static friction between the floor and the plank to prevent slipping is 0.5.

Step-by-step explanation:

(a) In this scenario, the forces that the corner and the floor exert on the plank can be determined by considering the equilibrium of forces. Since the plank is in equilibrium, the sum of the forces in the vertical direction must be zero. This means that the weight of the plank (200.0 N) must be balanced by the vertical components of the forces exerted by the corner and the floor. Therefore, both the corner and the floor exert a force of 200.0 N on the plank in the upward direction.

(b) To prevent the plank from slipping, the minimum coefficient of static friction between the floor and the plank can be found by considering the equilibrium of moments. The torque exerted by the weight of the plank at the point of contact with the floor must be balanced by the frictional torque. The torque can be calculated as the product of the weight of the plank and the distance from the point of contact to the center of mass. Therefore, the torque is (200.0 N)(2.0 m) = 400.0 N*m. The frictional torque can be calculated as the product of the frictional force and the distance from the point of contact to the center of mass. Therefore, the frictional torque is (f)(4.0 m), where f is the frictional force. Since the plank is not slipping, the frictional force can be determined using the coefficient of static friction, f = µsN, where N is the normal force exerted by the floor on the plank. Since the normal force is equal to the weight of the plank, the frictional force is µs(200.0 N). Equating the torque exerted by the weight of the plank to the frictional torque, we have (200.0 N)(2.0 m) = (µs(200.0 N))(4.0 m). Solving for µs, we find that the minimum coefficient of static friction is 0.5.