What Does the Scale Read Physics Massless Rod

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A bolt connecting the principal and rear frame of a mountain bike requires a torque of to tighten. If you are capable of applying of strength to a wrench in whatsoever given direction, what is the minimum length of the wrench that will issue in the required torque?

Correct answer:

Explanation:

The minimum length of the wrench will assume that the maximum force is practical at an angle of. Therefore, we can use the simplified expression for torque:

Here,  is the length of the wrench.

Rearranging for length and plugging in our values, we get:

A compatible rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left stop of the rod should a 0.6kg mass exist hung to balance the rod?

Possible Answers:

The rod cannot be balanced with this mass

50cm

42cm

48cm

45cm

Explanation:

The counterclockwise and clockwise torques about the pin point must be equal for the rod to balance. Taking the fulcrum as the pivot betoken, the counterclockwise torque is due to the rod'south weight, gravitational strength interim downward at the heart of the rod. If we use the pivot as our reference, and then the center of the rod is 15cm from the reference.

Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.

.

Solving for r gives r = 0.05m to the correct of the pivot, so forty + v cm from the left end of the rod.

A 2kg mass is suspended on a rope that wraps effectually a frictionless pulley fastened to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is fastened to a massless suspended platform, upon which 0.5kg weights may exist placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised past pulling on the rope.

What is the torque on the caster when the system is motionless?

Possible Answers:

19.6N*chiliad

9.8N*grand

0N*grand

10N*m

Explanation:

The net torque on the pulley is cipher. Remember that , assuming the forcefulness acts perpendicular to the radius. Because the caster is symmetrical in this problem (meaning the r is the same) and the tension throughout the unabridged rope is the aforementioned (pregnant F is the same), we know that the counterclockwise torque cancels out the clockwise torque, thus, the internet torque is zip.

In the image below, T_ane (due to the platform with the iv 0.5kg weights) = T_ii (the 2kg mass).

Ii students are balancing on a 10m seesaw. The seesaw is designed and so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the heart. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the lath that makes the seesaw is massless.

What distance from the center should the pupil on the right exist if they want the seesaw to stay parallel to the ground?

Correct answer:

Explanation:

Torque is defined by the equation. Since both students will exert a downwardly strength perpendicular to the length of the seesaw, . In our example, force is the force of gravity, given below, and is the distance from the center of the seesaw.

Since the torque must be naught in order for the seesaw to stay parallel (non motility), the lighter educatee on the correct must make his torque on the correct equal to the torque of the pupil on the left. Nosotros can determine the required distance by setting their torques equal to each other.

Two students are balancing on a 10m seesaw. The seesaw is designed and so that each side of the seesaw is 5m long. The pupil on the left weighs 60kg and is standing three meters away from the eye. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the lath that makes the seesaw is massless.

Imagine that the 2 students are sitting on the seesaw so that the torque is . Which of the post-obit changes will change the torque of the seesaw?

Possible Answers:

Both students move toward the center by i meter.

2 more than students become on the seesaw, each weighing 45kg. They both sit down on opposite ends of the seesaw, five meters away from the eye.

Another student stands perfectly on the center of the seesaw.

The heavier student moves forward 1m, while the lighter educatee moves forwards one.33m.

Correct answer:

Both students motion toward the center by i meter.

Caption:

Torque, in this example, is dependent on both the force exerted past the students also every bit their distances from the betoken of rotation. Equally a event, both students moving forwards by one meter will cause a nonzero torque on the seesaw. This is because the heavier pupil's ratio of strength and distance will upshot in less torque on his side than the lighter pupil.

A 3m axle of negligible weight is balancing in equilibrium with a fulcrum placed 1m from it's left end. If a force of 50N is practical on it's right terminate, how much force would needs to be applied to the left end?

Correct answer:

Explanation:

This a an example of rotational equilibrium involving torque. The formula for torque is , where is the angle that the force vector makes with the object in equilibrium and is the distance from the fulcrum to the point of the forcefulness vector. To accomplish equilibrium, our torques must exist equal.

Since the forces are applied perpendicular to the axle, becomes i. The distance of the fulcrum from the left end is 1m and its distance from the right end is 2m.

Since the 50N strength is twice as far from the fulcrum as the force that must be applied on the left side, it must be one-half equally strong as the strength on the left. The force on the left can be found to be 100N.

One side of a seesaw carries a mass iv meters from the fulcrum and a mass two meters from the fulcrum. To balance the seesaw, what mass should be placed nine meters from the fulcrum on the side contrary the showtime ii masses?

Right answer:

Caption:

For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides.

The full torque must be equal on both sides in club for the net torque to be cypher.

Substitute the formula for torque into this equation.

Now we tin can utilize the given values to solve for the missing mass.

The dispatch form gravity cancels from each term.

Ii masses hang beneath a massless meter stick. Mass ane is located at the 10cm mark with a weight of 15kg, while mass 2 is located at the 60cm mark with a weight of 30kg. At what betoken in between the two masses must the string be attached in order to balance the system?

Correct reply:

Explanation:

This trouble deals with torque and equilibrium. Noting that the cord is between the two masses we can utilise the torque equation of. We can apply the equation to find the torque. Since force is perpendicular to the distance nosotros can use the equation  (sine of ninety^o is 1). Force presented in this situation is gravity, therefore F=mg, and using the variable 10 equally a placement for the string nosotros can discover r.

x=43, thus the string is placed at the 43cm mark.

An attraction at a science museum helps teach students about the ability of torque. In that location is a long metal axle that has 1 pivot betoken. At ane end of the bar hangs a total sedan, and on the other end is a rope at which students can pull down, raising the car off the ground.

The beam is 40 meters long and the pivot signal is 5 meters from one end. A car of mass 500kg hangs from the brusk end of the beam. Neglecting the mass of the beam, what is the minimum mass of a student who tin can hang from the rope and brainstorm to raise the motorcar off the ground?

Possible Answers:

More data is needed to reply

Correct answer:

Explanation:

We are trying to notice what strength needs to be applied to the rope to result in a net of zero torque on the beam.

Torque applied by the car:

We tin use this to find the mass of a student that volition create the same corporeality of torque while hanging from the rope:

There is a  weight  to the left the center of a seesaw. What distance from the eye on the right side of the seesaw should Bob sit down so that the seesaw is counterbalanced?

Bob's mass is

Correct answer:

Explanation:

Torque is defined equally . In this case, is zero considering Bob and the weight are sitting directly on top of the seesaw; all of their weight is projected directly downward. Therefore, the torque that the weight applies is:

In order for the seesaw to balance, the torque practical by Bob must exist equal to .

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