Question

Exploration 2.2 For this Exploration, find a door at home. First apply a force perpendicular to a line through the pivot point (the door hinge) and the point of application. In other words, apply a force perpendicular to the door. Depending on the mass of the door, the force applied may not need to be very large. Just make sure you use the same force for all portions of this exercise. Then apply the force at an angle. Is the same force needed each time? Try applying a force parallel to a line through the pivot point and the point of application. Try the exercise again using different forces, i.e., push harder or lighter depending on angle. How do the forces and the angle affect how easy it is to close the door? Explain. The cross product of the force and the distance from the pivot is called the torque, t. Torque is a vector measured in units mN (meter Newtons). Mathematically, t=rxF The magnitude of the torque is t=rFsino = r F where is the angle between r, a vector drawn from the pivot point to the point of application of the force, and F.

Answer 1

axis of rotution Line through the pirot point & point of application of Force Point of pivot point of application of force Arsinex line of action of force where, r- is the distance between point of pivot f point of application of tone o- is the angle between r& line of action of force rsino = moment arm = it is the perpe- ndicular distance between axis of rotation & line of action of force OY ning the kihen, we close open a door then it states about an axis passing through its point of pirot. so for closing or ope- door (or rotation of door), torque must be applied on it because torque produces purely notational motion & mathe matically it is expressed as, Z= 8x² in vector form & c = rfsino- in magnitude form. Now, 1) Effect of angle on closing (rotation) of door: As given, it force is perpendicular to the door then line of action of force is also perpendicular to the door, then we have o go then, ? = rf singo=ff, this is a maximum torque acting on a door Tmax = 8F Now, this angle goes on decreasing means direction of application of force is not present perpendicular to door then' o decreases means torque decreases, when we apply force parallel to door then o=o or o=180° then we have, T= rfsinor Esin 180= ONm this is a minimum torque acting on (Note: At every point of application; magnitude of fore remains constant. (in this case)). a door.

Hence, when the direction or application from of force changesa perpendicular to parallel with respect to plane of door then angle '0' changes from go to o (or go to 180) then due to the change in angle magnitude of torque also changes. For go (or 270) magnitude of torque is maximum, hence in that case it becomes easy to close the door. But when angle is oo (or 180) then torque is zero then it is impossible to close the door, because no torque means no rotational motion. This is the effect of angle 2) Effect of force Now, it we are applixing force perpe- ndicular to the plane of door (ieyo go) & this direction of application of force does not changes or remains constant al coays If we changes point of application then, a) If point of application of force is present away from point of pirot or present at other end then we have T = Ersino, in this case & is maximum so torque is also maximum. =go, 8=&max - Tmax= Formax b) How, when point of application of fone is moved toward point of pirot then r goes on decreasing then due to which torque also decreases. When this point is at point of pivot then ,8=0&a=0Nm, means that it is impolsable to rotate be (or close) a door. Hence door eatily closed cohen point of application of fone is away from point of pivot. then point of application of force move toward point of pirot then & decre- ases due to cwhich torque decreases therefore in order to close a door we need to apply door to increase a torque means magnitude of force should be increased moves when point of application of fone toward point of pirot. can extra force on