6834 Lecture Notes - Lecture 4: Ground Reaction Force, Potential Energy, Kinetic Energy

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rosesnake902

21 Jun 2018

School

University of Canberra

Department

University of Canberra

Course

6834

Professor

all

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Lecture 4

Forces and linear kinetics

- Force

oAn interaction that causes a change in motion of an object

oForce is a vector (has magnitude and direction)

oInternal – muscle/bone

Tensile/compressive/shear forces

oExternal

Gravity (-9.81m/s)

Impacts weight as it is mass x gravity

Contact: once we start hitting objects

- Ground reaction force – only vertical

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Force over time

These two are the same

Acceleration and force are the same

Numbers will be different

Impart most forces in deepest level ie bottom of squat

oNet force (resultant)

Ultimate direction and magnitude

Co-linear forces

Forces acting in the same direction

Concurrent forces

oIf one is bigger - movement

oFree body diagrams

Take pictures; use arrows to describe where you think forces are

oStatic analysis

Equilibrium

Sum of all forces on body = 0

If not 0 = unbalanced = movement

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Document Summary

Force: an interaction that causes a change in motion of an object, force is a vector (has magnitude and direction, internal muscle/bone. Impacts weight as it is mass x gravity. Ground reaction force only vertical o o. Impart most forces in deepest level ie bottom of squat: net force (resultant) Concurrent forces: if one is bigger - movement, free body diagrams. Take pictures; use arrows to describe where you think forces are: static analysis. Sum of all forces on body = 0. If not 0 = unbalanced = movement o. Overcomes larger amount of friction between static and dynamic (when its moving) Horizontal; direction: green: breaking force, then we have compulsive force. Vertical: impact peak; little jolt in green, if wearing trainers; no little impact peak, blue; loading phase, big dip in blue; knee flexion little period of weightlessness, yellow; force starts to raise again blue/yellow to push off.

Related Questions

A rectangular loop of wire has dimensions of 0.500 m by 0.300 m. The loop is pivoted at the x-axis and lies in the XY plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop.

(a) What is the direction of the magnetic force exerted on wire segment ab?

(b) What is the direction of the torque associated with this force about an axis through the origin?

(d) What is the direction of the torque associated with this force about an axis through the origin?

(e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x-axis?

(f) Can they affect the motion of the loop in any way? Explain.

(g) What is the direction of the magnetic force exerted on segment bc?

(h) What is the direction of the torque associated with this force about an axis through the origin?

(i) What is the torque on segment ad about an axis through magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? O) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x-axis? (k) Compute the magnitude of the magnetic moment of the loop. (1) What is the angle between the magnetic moment vector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (1).

scarletbeetle53

1.

While helping a friend move, you push a 11.5 kg dolly with a forceof 11.0 N. Assume there are no other forces acting on the dolly. Ifthe dolly starts at rest, how far does it move in 3.50 s?
1 m

2.

At a lumberjack competition an axe with a mass of 1.25 kg is thrownat a target with a speed of 41 m/s. When it hits the target, itpenetrates to a depth of 0.055 m.
(a) What was the average force exerted by the target on theaxe?
1 N

(b) If the mass of the axe is halved, and the force exerted by thetarget on the axe remains the same, by what multiplicative factordoes the penetration depth change?
2
It doubles. It increases by a factor of the square root of two. Itdoes not change. It decreases by a factor of the square root oftwo. It halves.
Your work in question(s) will also be submitted or saved.
3

Your 1500 kg car pulls a 540 kg trailer away from a stop sign withan acceleration of 2.00 m/s2. (Note the answers should be given inkN.)
(a) What is the net force exerted by the car on the trailer?
1 kN

(b) What force does the trailer exert on the car?
2 kN

(c) What is the net force acting on the car?
3 kN

4.

Two boxes sit next to, and touching, each other on a frictionlesshorizontal surface. The lighter box has a mass of 5.31 kg and theheavier box has a mass of 7.29 kg.
(a) Find the contact force between these boxes when a horizontalforce of 6.1 N is applied to the light one.
1 N

(b) If the 6.1 N force is applied to the heavy box instead, is thecontact force between the boxes the same as, greater than, or lessthan the contact force in part (a)?
2

(c) Verify your answer to part (b) by calculating the contact forcein this case.
3 N

5.

A child sits in a wagon and together they have a mass of 25 kg. Thewagon is pulled with a constant speed by a handle that is at anangle of 22Â° above the horizontal. If the normal force exerted onthe wagon is 210 N, what is the force F applied to thehandle?
1 N

6.

An object acted on by three forces moves with constant velocity.One force acting on the object is in the positive x direction andhas a magnitude of 6.7 N; a second force has a magnitude of 4.5 Nand points in the negative y direction. Find the direction andmagnitude of the third force acting on the object.
1 N
2Â° (counterclockwise from the +x axis)

7.

A chain is used to lift a 1200 kg car vertically upward with anacceleration of 0.90 m/s2. What is the tension in the chain?
1 N

8.

A 22.5 kg box (m1) rests on a table.

(a) What is the weight of the box?
1 N
What is the normal force acting on it?
2 N (upward)
(b) A 12.5 kg box (m2) is placed directly on top of the 22.5 kgbox.
Determine the normal force that the table exerts on the 22.5 kgbox.
3 N (upward)
Determine the normal force that the 22.5 kg box exerts on the 12.5kg box.
4 N (upward)

9.

Your friend has a weight of 127 lb when standing on a scale. Youtake the scale to an elevator where you notice your friend's weighton the scale reads 108 lb. What are the direction and magnitude ofthe elevator's acceleration?
Direction 1
Magnitude 2 m/s2

10.

When you lift a package with a force of 82 N, the packageaccelerates upward at a rate of a. If, instead, you lift with aforce of 93 N, the package now has an acceleration of 8 multipliedby a.
(a) Find the weight of the package.
1 N

(b) Find the acceleration a.
2 m/s2

11.

One 3.1 kg paint bucket is hanging by a massless cord from another3.1 kg paint bucket, also hanging by a massless cord, as shownbelow.

(a) If the buckets are at rest, what is the tension in eachcord?
(lower cord) 1 N
(upper cord) 2 N

(b) If the two buckets are pulled upward with an acceleration of1.30 m/s2 by the upper cord, calculate the tension in eachcord.
(lower cord) 3 N
(upper cord)5

12)
Your friend has a pair of fuzzy dice with a mass of 1.32 g hangingfrom his rear-view mirror.
(a) When he accelerates from a stoplight, the dice deflect backwardtoward the rear of the car hanging at an angle of 6.18Â° relative tothe vertical. Find the tension in the string holding thedice.
N

(b) At what angle to the vertical will the tension in the string betwice the weight of the dice?
Â°

How fast can a racecar travel through a circular tum without skidding and hitting the wall? The answer could depend on several factors:

The weight of the car;
The friction between the tires and the road;
The radius of the circle;
The “steepness” of the turn.

In this project, we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.

A car of mass m moves with constant angular speed to around a circular curve of radius R (Figure 3.20). The curve is banked at an angle . If the height of the car off the ground is h, then the position of the car at time t is given by the function

As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure 3.21). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that for some positive constant . The constant is called the coefficient of friction.

Let denote the maximum speed the car can attain through the curve without skidding. In other words, is the fastest speed at which the car can navigate the turn. When the car is navigate at this speed, the magnitude of the centripetal force is

The next three questions deal with developing a formula that relates the speed to the banking angle .

Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.

The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure 3.22. Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately with a banking angle of . If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of . (The track has variable angle banking.)

The coefficient of friction for a normal fire in Elly conditions is approximately . Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately .

Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.

11. Suppose the measured speed of a car going along the outside edge of the turn is . Estimate the coefficient of friction for the car’s tires.

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6834 Lecture Notes - Lecture 4: Ground Reaction Force, Potential Energy, Kinetic Energy

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