Back To CourseAP Physics 1: Exam Prep
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Have you ever played a prank on someone, where they stand on a scale and you are behind them, and stealthily putting your foot on the scale to make them seem heavier? Your adding an extra force, in addition to the person's weight to the scale, and the scale responds by pushing back with the same force. This pushing back force is called the normal force, or contact force. Normal forces don't exist unless objects are touching and are always perpendicular to the surface(s). There is no general equation to calculate the normal force but there is a method to calculate the normal force. Let's get busy figuring this out.
Whenever dealing with forces, two major tools are required to analyze the situation: a free body diagram and Newton's second law, which is:
ΣF = ma (read as the net force equals mass times acceleration)
Let's look at a few situations where a mass is on a flat surface.
In the first situation, a construction crane has lifted a 500kg crate full of construction equipment, which is on a platform, connected to the crane's cable. The mass is at rest (500kg); what is the normal force exerted by the platform on the mass? Let's pull the mass out of the scenario and draw a free body diagram including the forces acting on it.
This is the free body diagram for when the mass is at rest:
mg is the weight of the crate; mg stands for mass x the acceleration due to gravity, which is weight, and N is the normal force provided by the platform on the crate. Now, we use Newton's second law (ΣF = ma) or the net force equals mass times acceleration.
For the vertical forces on the mass, giving us:
The normal force minus the weight of the crate equals zero.
Since there is no acceleration, we can solve for N in one step, resulting in:
N = mg = (500kg) (9.8 m/S^2) = 4,900 newtons
In the second situation, the construction worker calls for a crane to lift the 500kg crate to the top of the building. The mass accelerates at 1m/s^2. What is the normal force exerted by the platform on the mass while it's accelerating?
The free body diagram is the same as for situation 1:
ΣF = ma
Filling out the left side of Newton's second law, we get:
N - mg = ma
Solving for N results in:
N = (500kg)(1 m/s^2) + (500kg)(9.8 m/s^2)
N = 5,400 newtons
When the crane gets to the top of the building, the construction workers realize the wrong crate was delivered. They tell the crane operator to take that crate back to the ground. The crane lowers the mass towards the ground at 1 m/s^2. What is the normal force exerted by the platform on the mass? The free body diagram is still the same for this scenario and so is the expression we used in situation 2. We have a different acceleration:
Entering the values, we get:
N = ma + mg
N = (500kg) (-1 m/s^2) + (500kg)(9.8 m/s^2)
N = 4,400 newtons
When the crate is on the ground, a fork lift pushes on it at a 45 degree angle with a force of 1,000 newtons to move it off of the platform. What is the normal force exerted by the platform on the mass? The free body diagram is different in this scenario:
f represents the applied force exerted by the crane.
We have to break down the applied force into its vertical component. Here is the new free body diagram:
Filling out the left side of Newton's second law, for this scenario, results in:
The normal force minus the weight of the crate minus the vertical downward force provided by the forklift
Plugging in the values, we get:
N = (500kg)(9.8 m/s^2)+(1000)(0.707)
N = 5,607 newtons
Reviewing all of the normal forces provided by the platform, we see the largest normal force is when the crane was pushing the crate off of the platform. From the platform's perspective, it had to support the weight of the crate and the vertical force pushing down on the crate by the forklift. When the crate was at rest, the platform only had to support the weight of the crate. When the crate was accelerating upward, the platform had to support more than the crate's weight. The downward acceleration of the crate resulted in a lower normal force.
Let's now look at the normal force on an object on an inclined plane.
Let's look at the scenario where a mass is on a frictionless incline. Notice, we tilted our XY axis, so the X axis is parallel to the incline and the Y axis is perpendicular to the incline. The normal force is countering the component of gravity acting perpendicular to the ramp. Let's break the gravitational force down into its two component forces.
Let's calculate the normal force on this mass, starting with Newton's second law, we get:
The normal force minus mass times gravity times the cosign of theta equals zero
Solving for the normal force results in:
The normal force equals mg cosign theta
If any other force exerted on the mass makes the mass seem heavier with respect to the surface, the normal force will increase by that amount. If any other force makes the mass seem lighter than it is with respect to the surface, the normal force will decrease by that amount.
The normal force is the counter force to a mass pushing against a surface. The normal force is always perpendicular to the surface. When solving problems involving normal forces, it's important to draw a free body diagram, including all of the forces acting on the mass. All of the forces acting along the axis the normal force acts along, are plugged into the left side of Newton's second law (ΣF = ma). The product of the mass and acceleration is plugged in on the ride side. The resulting equation can be solved for the normal force.
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Back To CourseAP Physics 1: Exam Prep
12 chapters | 136 lessons