A 5.4 kg bag of groceries is in equilibrium on an incline of angle θ= 150. Find the magnitude of the normal force on the bag
To find the magnitude of the normal force on a bag of groceries on an incline, you can use the following equation:
N = mg * cos(θ)
Where:
N is the magnitude of the normal force
m is the mass of the bag (5.4 kg in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
θ is the angle of the incline (150 degrees)
Let's calculate it step-by-step:
Step 1: Convert the angle from degrees to radians.
θ = 150 degrees * (π/180) = 5π/6 radians
Step 2: Plug in the values into the equation.
N = (5.4 kg) * (9.8 m/s^2) * cos(5π/6)
Step 3: Calculate cos(5π/6) using a calculator.
cos(5π/6) is approximately -0.866
Step 4: Substitute the value of cos(5π/6) into the equation.
N ≈ (5.4 kg) * (9.8 m/s^2) * (-0.866)
Step 5: Perform the calculation.
N ≈ -47.64 N
Therefore, the magnitude of the normal force on the bag of groceries is approximately 47.64 N. Note that the negative sign indicates that the direction of the normal force is opposite to the gravitational force.
To find the magnitude of the normal force on the bag, we need to consider the forces acting on the bag in the y-direction (perpendicular to the incline).
In equilibrium, the sum of the forces in the y-direction is zero. We can break down the forces into their respective components:
1. Weight: The weight of the bag acts vertically downward with a magnitude equal to the mass of the bag (m = 5.4 kg) multiplied by the acceleration due to gravity (g = 9.8 m/s²): W = m * g. Breaking down the weight into its components, we have W_y = -m * g * cos(θ), where θ is the angle of the incline.
2. Normal force: The normal force is the force exerted by the incline on the bag perpendicular to the incline. Since the bag is in equilibrium, the normal force balances the component of the weight acting along the incline. Therefore, the normal force has a magnitude equal to the component of the weight perpendicular to the incline, which is given by N = m * g * sin(θ).
Since the bag is on an incline with a θ value of 150 degrees, we can calculate the magnitude of the normal force using the given information:
θ = 150 degrees
m = 5.4 kg
g = 9.8 m/s²
First, we need to convert the angle from degrees to radians by using the formula: θ_radians = θ_degrees * π/180.
θ_radians = 150 * π/180 = 5π/6 radians.
Now we can calculate the magnitude of the normal force:
N = m * g * sin(θ_radians)
N = 5.4 kg * 9.8 m/s² * sin(5π/6)
N ≈ 5.4 kg * 9.8 m/s² * 0.866 (since sin(5π/6) = 0.866)
N ≈ 47.394 N
Therefore, the magnitude of the normal force on the bag is approximately 47.394 Newtons.
How can an incline be of an angle greater than 90 degrees? Incline angles are measured to the horizontal, I thought.
Maybe you mean 15 degrees.
normal force= mg*Cos15