Newton’s Laws

Straight Line Motion A person pulls on a crate that weighs 600 N across a flat frictionless surface with a force of 30N. What is the motion of the crate? 1. Redraw the object of interest and label all forces (contact and field). 2. Apply Newton’s 1st Law to vertical case – since object has no motion in the vertical sense the forces in the vertical direction MUST add up to be zero.

Horizontally There is only one force in the horizontal direction so it must also be the net force So…

Force at an Angle 30o A person pulls on a crate that weighs 600 N across a flat frictionless surface with a force of 30N. What is the motion of the crate? Resolve Forces that are at an angle to coordinate system 30o Horizontally: Vertically:

Motion on a Ramp The person is pulling the same crate up a 25o ramp with a force of 300N. What is the motion of the crate? 25o Draw a free body diagram that labels direction of forces (contact and field) 25o +x +y We can anticipate the direction of the motion of the crate and simplify things by selecting a coordinate system in that direction. Notice that the weight vector is at an angle WRT our coordinate system. We’ll need to determine its components.

Ramp Problem 25o +y +x 2. Apply Newton’s 1st Law to vertical case – since the object has no motion in the vertical sense the forces in the vertical direction MUST add up to be zero. 3. There are two forces in the horizontal direction so we must find the net force

Motion on a Ramp The person is pulling the same crate up a 25o ramp when the rope breaks and the box begins to slide back down the ramp. What is the motion of the crate? 25o Draw a free body diagram that labels direction of forces (contact and field) Notice that the weight vector is at an angle WRT our coordinate system. We’ll need to determine its components. 25o +y +x

Ramp Example cont… +y +x 2. Apply Newton’s 1st Law to vertical case – since the object has no motion in the vertical sense the forces in the vertical direction MUST add up to be zero. 3. There is only one force in the horizontal direction so it must also be the net force

Atwood’s Machine 20 kg 10 kg Two masses hang over a pulley by means of a massless rope. Determine the motion of the system. Conceptual approach: Clearly the 20 kg object is falling and the 10 kg object is rising. If the 20kg object were only a 10 kg object then the system would be in equilibrium. So the effect is that there is an extra 10kg hanging on the left side. This provides a net force of 10(9.8) = 98 N This 98 N force acts on both masses (they both move due to this extra mass)

Atwood’s Machine cont… 20 kg 10 kg What is the tension in the rope? Conceptual approach: The entire system is accelerating at the rate of 3.26m/s2 On the 10kg object, the tension force provided by the string must be larger than the 10kg object’s weight. Newton’s 2nd Law says:

Friction

Friction as a Contact Force As the two surfaces are pushed closer together, more of the nooks and crannies come into contact and the friction increases. Friction arises from the roughness between two surfaces and how hard the two surface are pressed together Friction ALWAYS opposes motion.

Friction Defined Mathematically The force of friction depends upon two things: The inherent roughness between the two surfaces (m) The force which presses the two surfaces together Ordinarily, you might be tempted to use the weight of the object as the force which presses the two surfaces together – and in many case you would be right. But not always! In this case the force that pushes the two surfaces together is only one component of the weight. But is still equal to the Normal Force Here the weight is equal to the Normal Force

2 Cases -Static and Kinetic Friction fs=F f fs,max   F In both cases ms,k is the ratio of two forces fs,k/N and is dimensionless

Straight Line Motion - Revisited A person pulls on a crate that weighs 600 N across a flat surface with a force of 30N. The coefficient of kinetic friction between the crate and the surface is 0.28. The coefficient of static friction between the surfaces is 0.35. What is the motion of the crate? 1. Redraw the object of interest and label all forces (contact and field). (pick “up” to be positive and the direction of F to be positive) 2. Apply Newton’s 1st Law to vertical case – since object has no motion in the vertical sense the forces in the vertical direction MUST add up to be zero.

Horizontally There are two forces in the horizontal direction. So, we must determine the net force So…

Minimum Force to Move Crate If the crate is initially at rest then the person must overcome the static friction force. If the person pulls with a force of exactly 210N then the box is right on the verge of moving but doesn’t move. He must pull with a force slightly higher than 210N. Once the crate begins to move, he need only match the force of kinetic friction in order to pull the crate at constant speed. If he pulls with a force larger than 168N, then the crate is accelerating instead of traveling at constant speed in agreement with Newton’s 2nd Law.

Bobsled A bobsled of mass 180 kg is traveling at 50 mph (~25 m/s) along an icy track. The coefficient of kinetic friction between the sled runners and the ice is 0.05. If the sled were to coast to a stop at the bottom of the run, how far would it coast? 25 m/s In the vertical sense we should be able to see that the Normal force is equal in magnitude to the Weight of the sled In the horizontal sense:

Bobsled A bobsled of mass 180 kg is traveling at 50 mph (~25 m/s) along an icy track. The coefficient of kinetic friction between the sled runners and the ice is 0.05. If the sled were to coast to a stop at the bottom of the run, how far would it coast? 25 m/s In the horizontal sense:

Bobsled What must the coefficient of friction be if the sled had to be stopped within 50m? 25 m/s

Block on the wall An 8 N horizontal force F pushes a block weighing 12.0 N against a vertical wall . The coefficient of static friction between the wall and the block is 0.69, and the coefficient of kinetic friction is 0.49. Assume that the block is not moving initially. Horizontally the forces add up to be zero (no motion in this direction) Therefore, N and F have the same magnitude but opposite directions Since the weight is larger than the static friction force then the block is sliding down the wall (kinetic).

Block on the wall An 8 N horizontal force F pushes a block weighing 12.0 N against a vertical wall . The coefficient of static friction between the wall and the block is 0.69, and the coefficient of kinetic friction is 0.49. Assume that the block is not moving initially. Vertically the forces don’t add up to be zero (acceleration)

Block on the wall (revisited) How much force must the person exert against the same block to hold it in place on the wall? The coefficient of static friction between the wall and the block is 0.69, and the coefficient of kinetic friction is 0.49. Assume that the block is not moving initially. Vertically the forces have to add up to be zero.