How Do Mechanics and Physics Solve Practical Problems?

In summary: If you still can't find the answer, then your teacher probably didn't teach you how to find the center of mass. Let me know and I'll show you how to do that. #4: The bowling ball is rolling and therefore has both translational and rotational kinetic energies. You need to use the conservation of energy to calculate the final velocity. Set the initial potential energy equal to the final kinetic energy. mgh = 1/2 mv^2 + 1/2 Iw^2 where v is the final translational velocity, I is the moment of inertia of the bowling ball (1/2 mr^2), and w is the final angular velocity. You'll need to find the initial potential energy
  • #1
woodworker101
19
0
1. If the torque required to loosen a nut is 40.0 mN, what miminum force must be exerted by a mechanic to the end of a 30.0 cm wrench to remove the nut?

2. A uniform bridge 20.0 m long and weighting 4.00 x 10^5 N is supported by pillars located 3.00 m from each end. if a 1.96 x 10^4 N car is parked 8.00 m from one end of the bridge, how much force does each pillar exert.

3. A person lifts a 950 N box by pushing it up a incline. If the person exerts a force of 350 N along the incline, what is the mechanical advantage of the incline?

4. A 35 kg bowling ball with a radius of 13 cm starts from rest at the top of an incline 3.5 m in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (Assume the ball is a uniform solid sphere.)
 
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  • #2
This is in the wrong section and you need to show us the work you've done so far.
 
  • #3
the following four questions were from my test that i recentally took and i miss the day that we went over them and didn;t get the answers and haven;t had to time to check with my teacher about them, so i need some help on what the answers would be. thanks.

1. If the torque required to loosen a nut is 40.0 mN, what miminum force must be exerted by a mechanic to the end of a 30.0 cm wrench to remove the nut?

2. A uniform bridge 20.0 m long and weighting 4.00 x 10^5 N is supported by pillars located 3.00 m from each end. if a 1.96 x 10^4 N car is parked 8.00 m from one end of the bridge, how much force does each pillar exert.

3. A person lifts a 950 N box by pushing it up a incline. If the person exerts a force of 350 N along the incline, what is the mechanical advantage of the incline?

4. A 35 kg bowling ball with a radius of 13 cm starts from rest at the top of an incline 3.5 m in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (Assume the ball is a uniform solid sphere.)
 
  • #4
Don't you have a book?
 
  • #5
no forgot it at school but brought folder home, for #1 i get 133.3 N, #2 i don't know how to get N_1 and N_2, #3 i was told to do 950/350 but i believe he is just looking for a word answer, and #4 i get to the end but i don't know how to solve for vf^2.
 
  • #6
2. A uniform bridge 20.0 m long and weighting 4.00 x 10^5 N is supported by pillars located 3.00 m from each end. if a 1.96 x 10^4 N car is parked 8.00 m from one end of the bridge, how much force does each pillar exert.

Step One, write down everything you know:

Bridge length =[itex] L = 20m[/itex]
Bridge weight = [tex] m_{bridge}g = 4 x 10^5N[/itex]
Car weight = [tex] m_{car}g = 1.96 x 10^4N[/itex]
Car distance = [itex] d = 8m[/itex]
Pillars = [itex] P_1\ and \ P_2 = 3m \ and\ 17m[/itex]

Step Two, Identify your unknowns:

Force on both pillars due to the bridge and car.

Step Three, Formulate a set of equations that describe the problem:

The system is in equilibrium since nothing is moving. This means the net force at any point is zero. The net torque at any point is also zero. There are two axes of rotation, the two pillars. We will pick the left pillar [itex] P_1[/itex] as our zero point:

[tex] F_{net} = T_{net} = 0 [/tex]

The system exerts torque on both pillars. Find the center of mass of the system (bridge and man) and find the torque that this CM applies on both pillars.

[tex] \tau_{net} = \tau(P_1) + \tau(P2) = (CM_{system})(CM_{dist}) [/tex]

Step Four , Solve the equations for your unknown:

Step Five , Plug in your numbers and get an answer:

Step Six , Answer the question:


Can you finish the last 4 steps?
 
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  • #8
i am still lost even with the help of the other post. i don;t know where cm's come from and how to get them.
 
  • #9
was my answer to #1 correct?
 
  • #10
#1 The torque required is 40mN, your answer would be correct if the torque required was 40N. Remember the prefix 'm' means milli, or 10^-3
 
  • #11
3. A person lifts a 950 N box by pushing it up a incline. If the person exerts a force of 350 N along the incline, what is the mechanical advantage of the incline?

[PLAIN]http://www.ee.nmt.edu/~tubesing/missionpossible/advantage.htm said:
[/PLAIN]
Inclined Plane
With the inclined plane you are doing two things. You're increasing the distance the load travels to do the same work (ignoring friction). You're also dividing the force on the object into vertical and horizontal components. We're going to exhibit a horizontal force on an object that is moving diagonally upwards, thereby creating a force vector. A force vector has both a horizontal and vertical component. We will use the following formula, where m=mechanical advantage:

m = (length of inclined surface) / (height of incline)

This is a fraction derived from the formula 1 / (sin a) where a = angle of incline.

By trig identities we know that (sin a) = (height of incline) / (length of inclined surface)

So 1 / (sin a) inverts that fraction and we end up with the formula above.

Note that a straight vertical lift makes this fraction = 1 which denotes a 1-to-1 relationship between force on object and vertical load, therefore where there's "no mechanical advantage" then the result is 1. Keep in mind that the length of inclined surface is the hypotenuse of the triangle created (the length of the actual surface the load travels on), it is not the length of the base of the triangle.

Is that enough for #3? For #2, You must have been taught how to find center of masses before dealing with problems like this.
 
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Related to How Do Mechanics and Physics Solve Practical Problems?

1. What is wrench torque and why is it important for removing nuts?

Wrench torque is the measure of the rotational force applied by a wrench to a nut. It is important for removing nuts because it determines the amount of force needed to loosen the nut without causing damage.

2. How do I determine the necessary wrench torque to remove a nut?

The necessary wrench torque to remove a nut can be determined by consulting a torque chart or by using a torque wrench. The torque required will depend on the size and type of nut, as well as the material it is attached to.

3. Can I use any wrench to remove a nut or do I need a specific type?

While any wrench can technically be used to remove a nut, it is recommended to use a torque wrench specifically designed for the job. These wrenches allow for precise control of the applied force and help prevent over-tightening or damage to the nut.

4. Is there a risk of over-tightening a nut when using a torque wrench?

Yes, there is a risk of over-tightening a nut when using a torque wrench if the user is not careful. It is important to follow the manufacturer's instructions and use the appropriate torque setting for the specific nut being removed.

5. Can I reuse a nut that has been removed with a torque wrench?

It is generally not recommended to reuse a nut that has been removed with a torque wrench. The repeated application of torque can cause the nut to become damaged or weaken, which can lead to failure and potential safety hazards.

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