A construction worker attempts to lift a uniform beam off the floor and raise it to a vertical position. The beam is 2.77 m long and weighs 472 N. At a certain instant the worker holds the beam momentarily at rest with one end a distance d = 1.44 m above the floor, as shown in Fig. 12-75, by exerting a force on the beam. (a) What is the magnitude of ? (b) What is the magnitude of the (net) force of the floor on the beam? (c) What is the minimum value the coefficient of static friction between beam and floor can have in order for the beam not to slip at this instant?
990 N
To solve this problem, we'll use the principle of equilibrium. At the instant when the worker holds the beam at rest, the beam is in equilibrium, so the sum of the forces and torques acting on it must equal zero.
(a) To find the magnitude of the force F exerted by the worker, we need to balance the torques around the pivot point. The weight of the beam creates a torque in the counterclockwise direction, so the magnitude of F must create an equal and opposite torque in the clockwise direction. We can use the torque equation:
torque = force × lever arm
The weight of the beam creates a torque of:
τ_weight = weight × lever arm = (472 N) × (2.77 m - 1.44 m)
Since the beam is at rest, the torque created by F must be equal in magnitude but opposite in direction:
τ_F = -τ_weight
Therefore:
force × lever arm = -weight × lever arm
force = -weight = -472 N
So, the magnitude of the force F is 472 N.
(b) The net force of the floor on the beam can be determined using Newton's second law. Since the beam is at rest, the net force acting on it must be zero:
net force = F_floor - weight = 0
Therefore:
F_floor = weight = 472 N
So, the magnitude of the net force of the floor on the beam is 472 N.
(c) To find the minimum value of the coefficient of static friction between the beam and the floor, we can consider the horizontal forces acting on the beam. The vertical forces, including weight and the normal force, cancel each other out when the beam is at rest.
The maximum horizontal force the static friction can provide without causing the beam to slip is given by:
static friction = coefficient of static friction × normal force
In this case, the maximum horizontal force is equal to the force F exerted by the worker:
force = static friction
Substituting the values, we have:
472 N = μ × normal force
Since the normal force is equal to the weight of the beam, we can rewrite the equation as:
472 N = μ × weight
Solving for the coefficient of static friction (μ), we get:
μ = 472 N / weight
μ = 472 N / 472 N
So, the minimum value of the coefficient of static friction between the beam and the floor is 1.0.
To find the answers to these questions, we can utilize the principles of equilibrium and Newton's laws of motion. Let's go through each question step by step:
(a) To determine the magnitude of the force (F) exerted by the worker, we need to consider the equilibrium of forces acting on the beam. Since the beam is momentarily at rest, the sum of the vertical forces must be zero.
In this case, there are two vertical forces acting on the beam: the weight of the beam (472 N) acting downwards, and the force exerted by the worker (F) acting upwards. Since the beam is in equilibrium, these two forces must balance each other:
F - 472 N = 0
Solving for F, we find:
F = 472 N
Therefore, the magnitude of the force exerted by the worker is 472 N.
(b) To determine the magnitude of the net force of the floor on the beam, we need to consider the equilibrium of forces acting horizontally on the beam. Since the beam is momentarily at rest, the sum of the horizontal forces must be zero.
In this case, there are no horizontal forces acting on the beam. Therefore, the magnitude of the net force of the floor on the beam is zero.
(c) To calculate the minimum coefficient of static friction (μ) between the beam and the floor to prevent slipping, we need to consider the torque equilibrium. The torque exerted by the weight of the beam is equal and opposite to the torque exerted by the frictional force.
The torque exerted by the weight (τ_weight) is given by:
τ_weight = weight × perpendicular distance = 472 N × (1.44 m)
The torque exerted by the frictional force (τ_friction) is given by:
τ_friction = frictional force × perpendicular distance
Since the beam is at rest, these two torques must balance each other:
τ_weight = τ_friction
Substituting the expression for torque and rearranging, we get:
weight × perpendicular distance = frictional force × perpendicular distance
The perpendicular distance cancels out from both sides, leaving us with:
weight = frictional force
Substituting the values, we find:
472 N = μ × normal force
The normal force (N) is the force exerted by the floor on the beam and is equal to the weight of the beam (472 N) since the beam is at rest.
Therefore, the minimum value the coefficient of static friction (μ) can have for the beam not to slip is:
μ = weight / normal force = 472 N / 472 N = 1
Hence, the minimum value of the coefficient of static friction is 1.