Public

Edited Invalid date

0

0

Quiz

Chapter 30: How to Draw Effective Free-Body Diagrams

- An object has three forces pulling on it.

- You can choose the axis based on the motion of the object.
- Break A and B into components.
- Add the x-components to the y-components.

- The importance of dealing with direction is emphasized by a sign in front of each component.

- The final column should be used as a starting point for the problem.
- Almost all of the components of acceleration will be zero.

- A mass hangs from an elastic spring.
- The amount of stretching is proportional to the applied weight when the mass is attached vertically to the spring.
- Hooke's law is used to measure static forces.

- If the negative is used, it means that the applied force is good and the spring will go back in the opposite direction.
- As a static situation, a given spring can be used as a scale to indicate weight or other applied forces.

- The existence of an acceleration in the direction of the net force is implied by the second law.
- The acceleration will be 1 meter per second squared if a 10-kilogram mass is acted on by a 10-newton force from rest.
- We can use a free-body diagram to analyze all forces and apply the second law of motion.
- We need to be careful in choosing an appropriate frame of reference and coordinate system.
- Natural coordinate systems can be used if the mass is sliding down an incline in a way that the x - axis is parallel to the incline.

- In order to resolve the force of gravity into components, we must first identify the angle in the geometry.
- The component of weight along the incline is given by the angle of the incline.

- The normal force is equal to 85 N.

- The so-called conical pendulum is a point mass moving around a circle supported by a string.
- This situation is shown in a movie.

- The magnitude of the weight is balanced by the upward component of the tension in the string.
- The centripetal force is provided by the inward component of tension.
- The centripetal force is not an additional force.
- It is a description of force.

- If we specify that the mass is less than 1.5 m, we can see that, vertically, T cos 45 is greater.
- If T is 138.6 N, this means that if mg is 98 N, then T is 138.6 N.

- We find that v is 3.8 m/s using our known information.
- The components of forces must be resolved along the axes of the coordinate system.

- The opposing sliding motion is caused by the contact force between the two surfaces.
- The peaks and valleys are rough, like a mountain range.
- If a spring balance is attached to a mass and then pulled, the reading of the force scale before the mass begins to move provides a measure of static friction.
- The velocity of movement will be constant once the mass is moving.
- The net force is zero, so the acceleration will be zero.
- The reading of the scale will show how fast the particles are moving.
- The reading will be less than the beginning reading.

- The applied load pushing the mass into the surface is related to the frictional force.
- The normal force is measuring the push into the surface.
- The coefficients of friction are linked to the normal force by the proportionality constant.
- There are two coefficients of friction, one for static and the other for kinetic.

- The coefficients of static and kinetic are different.

- A mass is being pulled along a horizontal surface by a string.

- The normal force is less than the weight of the mass because of the upward component of the tension.

- These types of problems will usually give all the parameters and ask for the resulting acceleration or will state the mass is being pulled with constant velocity, allowing you to solve for a different unknown.

- A mass is held at rest at the top of an incline that makes a 40 degree angle with the ground.
- The incline is 1.5 m long.
- When the mass is released, it slides down the incline.

- We know that F is for f and k is for k.

- The object starts from rest.

- There are pushes and pulls that can be represented.

- The tendency of a mass to resist a force changing its state of motion is called inertia.

- The normal force is a force that moves away from a surface.

- Friction is a force that affects relative motion.

- In the absence of a net force, an object keeps its constant velocity.

- There is an equal but opposite reaction force for every action force.

- Earth's gravity pulls on a mass and causes weight to be directed downward.

- A frame of reference is moving at a constant rate.

- Free-body diagrams help in the analysis of forces by identifying and labeling all forces acting on a mass freed from the confines of the illustrated situation.

- The centripetal force is the net force that is directed inward toward the center.
- A pseudo force is a fake force attached to a mass that is moving.

- The proper free-body diagram is the key to solving force problems.
- Once a frame of reference and a coordinate system have been chosen, you need to resolve all forces into components.

- You can choose a coordinate system.
- If known, direct one axis in the direction of acceleration.

- If no sketch is provided, make one.

- A free-body diagram is needed for the situation.

- The chosen coordinate system can be used to resolve forces into components.

- The sum of all forces in a given direction is the second law.
- Set the summation to zero if the problem is static.
- If the situation is not static, set the summation to ma.
- Only applied forces are included in the diagram.

- It is always in a straight line.

- The centripetal force is always directed towards the circular path and parallel to the plane of the circle.
- The direction of gravity is always downward.

- Use the techniques for simultaneous equations to solve your equations.

- Positive and negative inward and outward forces are related.
- At the top and bottom of the roller coaster loop, compare the expressions for net force.

- M and m are hung over a massless pulley.

- At the bottom of the circle, the mass is observed to have a speed of 10 m/s.

- A car and driver have a lot of weight.
- The car goes over the top of the hill at a speed of 5 m/s.

- A hockey puck with a mass of less than one kilo is sliding along the ice.
- The puck's speed is 20 m/s.
- The floor has a 0.35 coefficient of friction, which is equal to the puck crossing over it.

- The natural length of the spring is 0.45 m and it is attached to the top of the incline.
- The incline is long.
- The spring is stretched down the incline because of a mass attached to it.

- The two blocks are being pushed along the floor by a 20-N force.

- A force of 20 N acts horizontally on a mass of 10 kg being pushed along a frictionless incline that makes a 30 degree angle with the horizontal, as shown below.

- A mass is released from rest on an incline that makes a 42 degree angle with the horizontal.

- A block is placed on top of another block.

- The block has a 0.35 coefficient of friction.
- There is a force horizontal F on the block.

- A road that is banked at an angle is not necessary for a car to stay on the road.
- A car is traveling at a speed of 25 m/s and the road has a radius of 40 m.

- The "rotor" is an amusement park ride that can be modeled as a rotating cylinder.
- A person is held against the sides of the ride as it rotates with a certain speed.
- The sides and the person have the same coefficients of static friction.

- We have to resolve the tensions into their x and y components.
- We have T 1 cos 45 and T 1 sin 45.
- The sum of all the forces must be zero.
- Since T 2 is completely horizontal, this means that T 1 cos 45 is T 2.
- The weight of T 1 must be balanced by the y component.

- The large mass is moving downward while the small mass is moving upward.
- The tension in the string is directed upward.

- We get ( M - m ) g /( M + m ) if we eliminate the tension T and solve for it.

- The downward force of gravity is opposed by the upward tension in the string.
- The upward net force must be achieved by doing a quick tip-to-tail graphical solution.

- The given values and solutions give us 35.7 N.

- The downward force of gravity is against this.
- The centripetal force F-c is produced by the combination.

- A sketch is shown.
- A downward net force must be achieved by doing a quick tip-to-tail graphical solution.
- 7,300 N upward is given by substituting values and solving for N.

- Net force is not acting on the puck when it is moving.
- The only net force that slows the puck down is the friction.

- The known numbers give a deceleration of -3.43 m/s 2.

- The distance the puck will travel before stopping is 58 m.

- The constant is 50 N/m.

- Hooke's law states that Fs is kx where x is the elongation in excess of the spring's natural length.
- The force is provided by the component of weight parallel to the incline.
- The answer is 0.646 m since this must be in excess of the spring's natural length.

- We drew a free-body diagram for the mass.

- The force that the 8 kilo mass exerts on the 2 kilo mass is represented by P. To find P, we use the known numbers.
- P is 16 N.

- The component of gravity is always given by the m g sin th.
- The force up the incline is at the same angle as the component parallel to it.

- The known numbers give a value of 3.17 m/s 2.

- The sum of all forces, in each direction, equals m a. T + 20 is 4 a, since the tension in the string will try to pull left.
- The tension pulls up against gravity.
- T - 18.6 is 2 a.
- We get T of 19.7 N by eliminating a.

- The known number gives us an acceleration of 0.67 m/s 2.
- On an incline, the normal force is given by mg cos th.
- The downward force of gravity against the incline was again opposed by this force.
- The two forces are combined in the downward direction.

The coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the

- N-1 is up and N-1 is down on the mass.

- The f-s is to the right on the 3- kilo mass and the left on the 7- kilo mass.

- The applied force, F, is acting on the 7- kilogram block.
- The force F must accelerate both blocks.

- Since we have horizontal motion, the normal force is equal to the combined weights, we can write.
- F is 84.3 N and we get it with Solving for F.

- The normal force on this surface is the weight of the block, as shown in the free-body diagram drawn in part a.

- From the vertical components, we can see that the N cos th is.

- We find that tan th is 1.59 and th is 57.8deg.

- The normal force is the same as the person and supplies the inward centripetal force.
- The person is slide down by Friction against the wall.
- To maintain equilibrium, the key is to be fast.

- The apparent weight of the mass at the equator is reduced by an upward force in the frame of reference.
- Only very sensitive scales can measure this effect.

- This is a very difficult question.
- The action and reaction pairs act on different objects.
- An object can move in a frame of reference if the applied force results in a net force.
- The force reaction can be experienced by a different object, which may or may not have its own acceleration.

Study Panel

Review flashcards and saved quizzes

Getting your flashcards

Review

Quizzes

Mine

Others

Notifications

U

Profile

Mobile App

Privacy & Terms

Feedback

Need Help?

Tutorial

Log out