Wrench torque to remove nut

[woodworker101]

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.)

 

[Pengwuino]

This is in the wrong section and you need to show us the work you've done so far.

 

[woodworker101]

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.)

 

[Pengwuino]

Don't you have a book?

 

[woodworker101]

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.

 

[whozum]

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 =$L = 20m$
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 = $d = 8m$
Pillars = $P_1\ and \ P_2 = 3m \ and\ 17m$

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 $P_1$ as our zero point:

$$F_{net} = T_{net} = 0$$

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.

$$\tau_{net} = \tau(P_1) + \tau(P2) = (CM_{system})(CM_{dist})$$

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?

 

[whozum]

If you still need help with #2 the latter half of this thread may be of help:

https://www.physicsforums.com/showthread.php?t=69413

 

[woodworker101]

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.

 

[woodworker101]

was my answer to #1 correct?

 

[whozum]

#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

 

[whozum]

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]
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.