A 1,080-N crate is being pushed across a level floor at a constant speed by a force of 290 N at an angle of 20.0° below the horizontal, as shown in the figure (a) below.
(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 290-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).
To solve these problems, we need to analyze the forces acting on the crate and apply Newton's second law.
(a) To find the coefficient of kinetic friction between the crate and the floor, we need to consider the forces acting on the crate. The force pushing the crate is given as 290 N at an angle of 20.0° below the horizontal.
Step 1: Resolve the force into horizontal and vertical components:
The horizontal component can be found using the equation: F_horizontal = F * cos(θ)
F_horizontal = 290 N * cos(20.0°) = 274.21 N
Step 2: Find the force of friction:
The force of friction can be found using the equation: F_friction = μ * N
Since the crate is moving at a constant speed, the frictional force must be equal in magnitude and opposite in direction to the horizontal component of the applied force.
F_friction = F_horizontal = 274.21 N
Step 3: Calculate the normal force:
The normal force is the force exerted by the floor on the crate and is equal to the weight of the crate. Therefore, N = mg, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the weight (mg) of the crate is 1,080 N, the normal force N = 1,080 N.
Step 4: Substitute the values into the equation for the coefficient of friction:
F_friction = μ * N
274.21 N = μ * 1,080 N
μ = 0.254 (rounded to three significant figures)
Therefore, the coefficient of kinetic friction between the crate and the floor is approximately 0.254.
(b) To find the acceleration of the crate when a force of 290 N is pulling it at an angle of 20.0° above the horizontal, we need to consider the forces acting on the crate.
Step 1: Resolve the force into horizontal and vertical components:
The horizontal component can be found using the equation: F_horizontal = F * cos(θ)
F_horizontal = 290 N * cos(20.0°) = 274.21 N
Step 2: Calculate the net force in the horizontal direction:
The net force in the horizontal direction is the difference between the horizontal component of the applied force and the force of friction.
Net force = F_horizontal - F_friction
Net force = 274.21 N - 274.21 N = 0 N
Step 3: Use Newton's second law to find the acceleration:
The net force is equal to the mass of the crate multiplied by the acceleration. Therefore, we have Net force = m * a.
Since the net force is zero, we can conclude that the acceleration of the crate is also zero.
Therefore, when the 290-N force is pulling the crate at an angle of 20.0° above the horizontal, the acceleration of the crate is zero.