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A bolt connecting the main and rear frame of a mountain bike requires a torque of to tighten. If you are capable of applying of force to a wrench in any given direction, what is the minimum length of the wrench that will result in the required torque? The minimum length of the wrench will assume that the maximum force is applied at an angle of. Therefore, we can use the simplified expression for torque:. Here, is the length of the wrench. A uniform rod of length 50cm and mass 0. At what distance from the left end of the rod should a 0.

The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. If we use the pivot as our reference, 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.

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0. The other end of the rope is attached to a massless suspended platform, upon which 0. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

The net torque on the pulley is zero. Remember thatassuming the force acts perpendicular to the radius. Because the pulley is symmetrical in this problem meaning the r is the same and the tension throughout the entire rope is the same meaning F is the samewe know that the counterclockwise torque cancels out the clockwise torque, thus, the net torque is zero. In the image below, T 1 due to the platform with the 4 0. Two students are balancing on a 10m seesaw. The seesaw is deed 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 center.

The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground? Torque is defined by the equation. Since both students will exert Some strings attached 45 Portland Maine area 45 downward force perpendicular to the length of the seesaw. In our case, force is the force of gravity, given below, and is the distance from the center of the seesaw.

Since the torque must be zero in order for the seesaw to stay parallel not movethe lighter student on the right must make his torque on the right equal to the torque of the student on the left. We can determine the required distance by setting their torques equal to each other. Imagine that the two students are sitting on the seesaw so that the torque is.

Which of the following changes will alter the torque of the seesaw? Two more students get on the seesaw, each weighing 45kg. They both sit on opposite ends of the seesaw, five meters away from the center. Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation.

As a result, both students moving forward by one meter will cause a nonzero torque on the seesaw. This is because the heavier student's ratio of force and distance will result in less torque on his side than the lighter student. A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from it's left end.

If a force of 50N is applied on it's right end, how much force would needs to be applied to the left end? This a an example of rotational equilibrium involving torque. The formula for torque iswhere 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 force vector. To achieve equilibrium, our torques must be equal. Since the forces are applied perpendicular to the beam, becomes 1. The distance of the fulcrum from the left end is 1m and its distance from the right end is Some strings attached 45 Portland Maine area 45.

Since the 50N force is twice as far from the fulcrum as the force that must be applied on the left side, it must be half as strong as the force on the left. The force on the left can be found to be N. One side of a seesaw carries a mass four 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 opposite the first two masses? For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides. Two masses hang below a massless meter stick. Mass 1 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 point in between the two masses must the string be attached in order to balance the system? This problem deals with torque and equilibrium.

Noting that the string is between the two masses we can use the torque equation of. We can use the equation to find the torque. Since force is perpendicular to the distance we can use the equation sine of 90 o is 1. An attraction at a science museum helps teach students about the power of torque. There is a long metal beam that has one pivot point. At one end of the bar hangs a full 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 point is 5 meters from one end. A car of mass kg hangs from the short end of the beam. Neglecting the mass of the beam, what is the minimum mass of a student who can hang from the rope and begin to raise the car off the ground?

We are trying to find what force needs to be applied to the rope to result in a net of zero torque on the beam. We can use this to find the mass of a student that will create the same amount of torque while hanging from the rope:. There is a weight to the left the center of a seesaw. What distance from the center on the right side of the seesaw should Bob sit so that the seesaw is balanced? Bob's mass is. Torque is defined as. In this case, is zero because 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 applied by Bob must be equal to. If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

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Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Hanley Rd, Suite St. Louis, MO Subject optional. Home Embed. address: Your name:. Example Question 1 : Torque. Possible Answers:. Correct answer:. Explanation : The minimum length of the wrench will assume that the maximum force is applied 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:. Report an Error.

Possible Answers: 42cm. Correct answer: 45cm. Explanation : The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. What is the torque on the pulley when the system is motionless? Explanation : The net torque on the pulley is zero. Explanation : Torque is defined by the equation. Example Question 2 : Torque. Possible Answers: Two more students get on the seesaw, each weighing 45kg. Both students move toward the center by one meter. The heavier student moves forward 1m, while the lighter student moves forward 1.

Another student stands perfectly on the center of the seesaw. Correct answer: Both students move toward the center by one meter. Explanation : Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation. Explanation : This a an example of rotational equilibrium involving torque. Explanation : For the seesaw to be balanced, the system must be in rotational equilibrium. The total torque must be equal on both sides in order for the net torque to be zero.

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*AP Physics 1 : Torque*