Ask question

Ask question

All categories

PilotLPTM [1.2K]

3 years ago

13

A bridge is supported by two piers located 14 meters apart. Both the left and right piers provide an upward force on the bridge,

labeled FL and FR respectively. (a) If a 1250 kg car comes to rest at a point 5 meters from the left pier, how much force (in N) will the bridge provide to the left and right piers? (Enter the magnitudes.)

1 answer:

almond37 [142]3 years ago

5 0

Answer:

4375 N, 7875 N

Explanation:

since the body is equilibrium

total upward force = total downward force

w weight of the bridge = FL + FR

when the car was introduced,

total downward force = total upward force

FL₁ + FR₁ = w + (m × acceleration due to gravity) with w = FL + FR

then FL₁ + FR₁ = FL + FR + Mcg

FL₁ + FR₁ - FL - FR = Mcg

ΔFL + ΔFR = 1250 × 9.8 = 12250 N

taken the left as the pivot point and using the principle of moment

ΔFR × 14 m = 12250 N × 5 m + (ΔFL × 0 m) since the left is the pivot point.

ΔFR = 61250 / 14 = 4375 N

but

ΔFL + ΔFR = 12250 N

ΔFL = 12250 - 4375 = 7875 N

You might be interested in

Suppose that the model presented by student 1 is correct. Based on the information provided, what would be the bond angle in a m

Ugo [173]

Question: Predicting the shape of a molecule is relatively straight forward. A molecule's shape will always be determined by the number of electron pairs around the central atom. The number of electron pair corresponds to the number of atoms that are bound to the central atom of the molecule. For example, water contains two hydrogen atom bound to one atom of oxygen, giving the molecule a linear geometry.

Suppose that the model presented by student 1 is correct. Based on the information provided, what would be the bond angle in a molecule of perchlorate ion.

Answer: Suppose that the model presented by student 1 is correct The (perchlorate ion) will be a tetrahedral shape, O-Cl-O bond angle 109.5 due to four groups of bonding electrons and no lone pairs of electrons.

8 0

3 years ago

If Fg=mg solve for g

Bas_tet [7]

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

6 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago

The bob of a pendulum swings back and forth with a total mechanical energy of 300 J. What is the kinetic energy of the bob when

zhenek [66]

at the lowest point in the trajector, the kinetic energy of the bob is 300 J.

Explanation:

The total mechanical energy of the bob at any point of its motion is given by

$E=KE+PE$

Where

$KE=\frac{1}{2}mv^2$ is the kinetic energy, where

m is the mass of the bob

v is its speed

$PE=mgh$ is the gravitational potential energy, where

g is the acceleration of gravity

h is the height of the bob, measured with respect to the lowest point of the trajector

In absence of friction, the total mechanical energy E remains constant. So we have:

- When the bob swings upward, the PE increases (because h increases) and the KE decreases (so the speed decreases). At the highest point in the trajector, the speed of the bob is zero (v=0), so its KE is also zero and all the mechanical energy is potential energy: U = 300 J

- When the bob swings downward, the PE decreases (because h decreases) and the KE increases (so the speed increases). At the lowest point in the trajectory, the height has become zero (h=0), so the PE is zero and all the mechanical energy is kinetic energy: KE = 300 J

Therefore, at the lowest point in the trajector, the kinetic energy of the bob is 300 J.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

4 0

3 years ago

As a rocket rises, its kinetic energy changes

Bond [772]

<span>Transformed into potential energy</span>

3 0

4 years ago

Read 2 more answers