PHYS 101 Chapter Notes -Net Force, Angular Velocity, Angular Acceleration

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rosewildebeest832

24 Mar 2013

School

Simon Fraser University

Department

Physics

Course

PHYS 101

Professor

Andrew Debenedictis

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Document Summary

First note that the system is in static equilibrium. Therefore, net torque and net force are 0. Draw the fbd and list the different forces. Torque for the weight of the beam + the torque for the person"s weight=torque of the. Using this torque formula, find the torque for each and equate them. Pay attention to the r for each one from the info in the question. Also note that the angle for the weight torques is the one between the beam closest to the pivot point (origin) and the weight force. It should indicate somethingsimilarly, the angle for the tension is the angle between the tension and the line of action extending from the radius. Hint the angle is theta+90 degree as indicated in the drawing. You are asked to find the horizontal and vertical pin forces or similarly, nx and ny. n=normal force.

Related Questions

In certain cities, parking spaces are at a premium and anyavailable surface will be used for parking, even highly slopedroofs. Consider the SUV parked in the picture above. If you are thedriver, how can you be sure that your car wonât flip overbackwards?
The schematic diagram below gives a simplified model of thesituation with the SUV on the slope. We are approximating the SUVas a uniform cube three meters long and two meters high, mounted ontires a half-meter in diameter that are set in a half meter fromthe edges of the vehicle. The SUV has a mass of 1500 kg. Forsimplicity, we will ignore the mass of the tires. Assume that thecoefficient of static friction between the tires and the road is0.8. Using the principles of static equilibrium, letâs determinewhether the car will stay on the slope. In order to maintain staticrotational equilibrium, the magnitudes of the normal forces on theforward and rear wheels must adjust to create a situation of zeronet torque. In order to maintain static translational equilibrium,the static friction forces between the tires and the road mustadjust to create a situation of zero net force. If the angle Î¸ getstoo big, we will have a problem...
1. First, letâs assume that the angle Î¸ is 30 degrees (note thewheels are locked and donât roll) and the car is not moving. Whatis the magnitude of the normal force from the road on the frontwheels (top of incline)? On the back wheels (bottom of incline)?These normal force are unequal- why?
2. Now letâs take the situation to the extreme: what is the maximumvalue of the angle Î¸ which is possible, if the SUV is to remain instatic equilibrium? First, determine the maximum angle that the carwonât slide, and then determine the maximum angle that the carwonât rotate. What happens to the SUV first if the angle isexceeded? What is the normal force on the front (top of incline)wheels in this situation?
3. Assume that it rains, and the coefficient of static friction isreduced from 0.8 to 0.5. Now what is the maximum value of the angleÎ¸ for which the car can remain in static equilibrium? If the angleÎ¸ exceeds this value, what happens to the SUV?

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°

PHYS 101 Chapter Notes -Net Force, Angular Velocity, Angular Acceleration

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