Different types of forces, Grade 11 physics Tension Tension is the magnitude of the force that exists in objects like ropes, chains and struts that are providing support. For example, there are tension forces in the ropes supporting a child’s swing hanging from a tree. Contact and non-contact forces In this chapter we have come across a number of different types of forces, for example a push or a pull, tension in a string, frictional forces and the normal. These are all examples of contact forces where there is a physical point of contact between applying the force and the object. Non-contact forces are forces that act over a distance, for example magnetic forces, electrostatic orces and gravitational forces. When an object is placed on a surface, two types of surface forces can be identified. Friction is a force that acts between the surface and the object and parallel to the surface. The normal force is a force that acts between the object and the surface and parallel to the surface. The normal force A 5 kg box is placed on a rough surface and a 10 N force is applied at an angle of 36,9◦ to the horizontal. The box does not move. The normal force (N or FN ) is the force between the box and the surface acting in the vertical direction. If this force is not present the box would fall through the surface because the force of gravity pulls it downwards. The normal force therefore acts upwards. We can calculate the normal force by considering all the forces in the vertical direction. All the forces in the vertical direction must add up to zero because there is no movement in the vertical direction. Figure 12.12: Friction and the normal forceThe most interesting and illustrative normal force question, that is often asked, has to do with a scale in a lift. Using Newton’s third law we can solve these problems quite easily. When you stand on a scale to measure your weight you are pulled down by gravity. There is no acceleration downwards because there is a reaction force we call the normal force acting upwards on you. This is the force that the scale would measure. If the gravitational force were less then the reading on the scale would be less. Worked Example 1: Normal Forces 1 Question: A man with a mass of 100 kg stands on a scale (measuring newtons). What is the reading on the scale? Hướng dẫn Step 1 : Identify what information is given and what is asked for We are given the mass of the man. We know the gravitational acceleration that acts on him is 9,8 = m·s$^{−2}$. Step 2 : Decide what equation to use to solve the problem The scale measures the normal force on the man. This is the force that balances gravity. We can use Newton’s laws to solve the problem: F$_{r}$ = F$_{g}$ + F$_{N}$ where Fr is the resultant force on the man. Step 3 : Firstly we determine the force on the man due to gravity F$_{g}$ = mg = (100 kg)(9,8 m · s$^{−2}$) = 980 kg · m · s$^{−2}$ = 980 N downwards Step 4 : Now determine the normal force acting upwards on the man We now know the gravitational force downwards. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 0 - so F$_{r}$ = 0. F$_{r}$ = F$_{g}$ + F$_{N}$ 0 = −980 N + F$_{N}$ F$_{N}$ = 980 N upwards Step 5 : Quote the final answer The normal force is 980 N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. The reading on the scale is 980 N. Now we are going to add things to exactly the same problem to show how things change slightly. We will now move to a lift moving at constant velocity. Remember if velocity is constant then acceleration is zero. Worked Example 2: Normal Forces 2 Question: A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that moving downwards at a constant velocity of 2 m·s$^{−1}$. What is the reading on the scale? Hướng dẫn Step 6 : Identify what information is given and what is asked for We are given the mass of the man and the acceleration of the lift. We know the gravitational acceleration that acts on him. Step 7 : Decide which equation to use to solve the problem Once again we can use Newton’s laws. We know that the sum of all the forces must equal the resultant force, F$_{r}$ . F$_{r}$ = F$_{g }$+ F$_{N}$ Step 8 : Determine the force due to gravity F$_{g}$ = mg = (100 kg)(9,8 m · s$^{−2}$) = 980 kg · m · s$^{−2}$ = 980 N downwards Step 9 : Now determine the normal force acting upwards on the man The scale measures this normal force, so once we have determined it we will know the reading on the scale. Because the lift is moving at constant velocity the overall resultant acceleration of the man on the scale is 0. If we write out the equation: F$_{r }$= F$_{g}$ + F$_{N}$ ma = F$_{g}$ + F$_{N}$ (100)(0) = −980 N + F$_{N}$ F$_{N }$= 980 N upwards Step 10 : Quote the final answer The normal force is 980 N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. The reading on the scale is 980 N. In the previous two examples we got exactly the same result because the net acceleration on the man was zero! If the lift is accelerating downwards things are slightly different and now we will get a more interesting answer! Worked Example 3: Normal Forces 3 Question: A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is accelerating downwards at 2 m·s$^{−2}$. What is the reading on the scale? Hướng dẫn Step 1 : Identify what information is given and what is asked for We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration also acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton’s laws. We know that the sum of all the forces must equal the resultant force, F$_{r}$ . F$_{r}$ = F$_{g}$ + F$_{N}$ Step 3 : Determine the force due to gravity, F$_{g}$ F$_{g}$ = mg = (100 kg)(9,8 m · s$^{−2}$) = 980 kg · m · s$^{−2}$ = 980 N downwards Step 4 : Determine the resultant force, F$_{r}$ The resultant force can be calculated by applying Newton’s Second Law: F$_{r}$ = ma = (100)($^{−2}$) = $^{−2}$00 N = 200 N down Step 5 : Determine the normal force, F$_{N}$ The sum of all the vertical forces is equal to the resultant force, therefore F$_{r }$= F$_{g}$ + F$_{N}$ $^{−2}$00 = −980 + F$_{N}$ F$_{N}$ = 780 N upwards Step 6 : Quote the final answer The normal force is 780 N upwards. It balances the gravitational force downwards just enough so that the man only accelerates downwards at 2 m·s$^{−2}$. The reading on the scale is 780 N. Worked Example 4: Normal Forces 4 Question: A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is accelerating upwards at 4 m·s$^{−2}$. What is the reading on the scale? Hướng dẫn Step 1 : Identify what information is given and what is asked for We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration also acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton’s laws. We know that the sum of all the forces must equal the resultant force, F$_{r}$ . F$_{r}$ = F$_{g}$ + FN Step 3 : Determine the force due to gravity, F$_{g}$ F$_{g}$ = mg = (100 kg)(9,8 m · s$^{−2}$) = 980 kg · m · s$^{−2}$ = 980 N downwards Step 4 : Determine the resultant force, F$_{r}$ The resultant force can be calculated by applying Newton’s Second Law: F$_{r}$ = ma = (100)(4) = 400 N upwards Step 5 : Determine the normal force, F$_{N}$ The sum of all the vertical forces is equal to the resultant force, therefore F$_{r}$ = F$_{g}$ + F$_{N}$ 400 = −980 + F$_{N}$ F$_{N}$ = 1380 N upwards Step 6 : Quote the final answer The normal force is 1380 N upwards. It balances the gravitational force downwards and then in addition applies sufficient force to accelerate the man upwards at 4m·s$^{−2}$. The reading on the scale is 1380 N. Xem thêm: Physics 11.II Force, Momentum, Impulse High School Students Studying the Sciences Physics