Newton’s Second Law of Motion, Object on an inclined plane, Lifts and rockets, Grade 11 physics Object on an inclined plane When we place an object on a slope the force of gravity (F$_{g}$ ) acts straight down and not perpendicular to the slope. Due to gravity pulling straight down, the object will tend to slide down the slope with a force equal to the horizontal component of the force of gravity (F$_{g}$ sin θ).The object will ’stick’ to the slope due to the frictional force between the object and the surface. As you increase the angle of the slope, the horizontal component will also increase until the frictional force is overcome and the object starts to slide down the slope. The force of gravity will also tend to push an object ’into’ the slope. This force is equal to the vertical component of the force of gravity (F$_{g}$ cos θ). There is no movement in this direction as this force is balanced by the slope pushing up against the object. This ?pushing force? is called the normal force (N) and is equal to the resultant force in the vertical direction, F$_{g}$ sin θ in this case, but opposite in direction. Important: Do not use the abbreviation W for weight as it is used to abbreviate ’work’. Rather use the force of gravity F$_{g}$ for weight. Worked Example 1: Newton II - Box on inclined plane Question: A body of mass M is at rest on an inclined plane. What is the magnitude of the frictional force acting on the body? A/ Mg B/ Mg cos θ C/ Mg sin θ D/ Mg tan θ Hướng dẫn Step 1 : Analyse the situation The question asks us to identify the frictional force. The body is said to be at rest on the plane, which means that it is not moving and therefore there is no resultant force. The frictional force must therefore be balanced by the force F up the inclined plane. Step 2 : Choose the correct answer The frictional force is equal to the horizontal component of the weight (Mg) which is equal to Mg sin θ. Worked Example 2: Newton II - Object on a slope Question: A force T = 312 N is required to keep a body at rest on a frictionless inclined plane which makes an angle of 35◦ with the horizontal. The forces acting on the body are shown. Calculate the magnitudes of forces P and R, giving your answers to three significant figures. Hướng dẫn Step 1 : Find the magnitude of P We are usually asked to find the magnitude of T, but in this case T is given and we are asked to find P. We can use the same equation. T is the force that balances the horizontal component of P (P$_{x}$) and therefore it has the same magnitude. T = P sin θ => 312 = P sin 35◦ => P = 544 N Step 2 : Find the magnitude of R R can also be determined with the use of trigonometric ratios. The tan or cos ratio can be used. We recommend that you use the tan ratio because it does not involve using the value for P (for in case you made a mistake in calculating P). tan 55° = R/T => R = 446N Note that the question asks that the answers be given to 3 significant figures. We therefore round 445,6 N up to 446 N. Lifts and rockets So far we have looked at objects being pulled or pushed across a surface, in other words horizontal motion. Here we only considered horizontal forces, but we can also lift objects up or let them fall. This is vertical motion where only vertical forces are being considered. Let us consider a 500 kg lift, with no passengers, hanging on a cable. The purpose of the cable is to pull the lift upwards so that it can reach the next floor or to let go a little so that it can move downwards to the floor below. We will look at five possible stages during the motion of the lift. Stage 1: The 500 kg lift is stationary at the second floor of a tall building. Because the lift is stationary (not moving) there is no resultant force acting on the lift. This means that the upward forces must be balanced by the downward forces. The only force acting down is the force of gravity which is equal to (500 x 9,8 = 4900 N) in this case. The cable must therefore pull upwards with a force of 4900 N to keep the lift stationary at this point. Stage 2: The lift moves upwards at an acceleration of 1 m·s$^{$^{−2}$}$. If the lift is accelerating, it means that there is a resultant force in the direction of the motion. This means that the force acting upwards is now bigger than the force of gravity F$_{g}$ (down). To find the magnitude of the force applied by the cable (F$_{c}$ ) we can do the following calculation: (Remember to choose a direction as positive. We have chosen upwards as positive.) F$_{R}$ = ma F$_{c}$ + F$_{g}$ = ma F$_{c}$ + (−4900) = (500)(1) F$_{c}$ = 5400 N upwards The answer makes sense as we need a bigger force upwards to cancel the effect of gravity as well as make the lift go faster. Stage 3: The lift moves at a constant velocity. When the lift moves at a constant velocity, it means that all the forces are balanced and that there is no resultant force. The acceleration is zero, therefore F$_{R}$ = 0. The force acting upwards is equal to the force acting downwards, therefore F$_{c}$ = 4900 N. Stage 4: The lift slow down at a rate of 2m·s$^{$^{−2}$}$. As the lift is now slowing down there is a resultant force downwards. This means that the force acting downwards is bigger than the force acting upwards. To find the magnitude of the force applied by the cable (F$_{c}$) we can do the following calculation: Again we have chosen upwards as positive, which means that the acceleration will be a negative number. F$_{R }$= ma F$_{c }$+ F$_{g}$ = ma F$_{c}$ + (−4900) = (500)($^{−2}$) F$_{c}$ = 3900 N upwards This makes sense as we need a smaller force upwards to ensure a resultant force down. The force of gravity is now bigger than the upward pull of the cable and the lift will slow down. Stage 5: The cable snaps. When the cable snaps, the force that used to be acting upwards is no longer present. The only force that is present would be the force of gravity. The lift will freefall and its acceleration can be calculated as follows: F$_{R}$ = ma F$_{c}$ + F$_{g}$ = ma 0 + (−4900) = (500)(a) a = −9,8 m · s$^{$^{−2}$}$ a = 9,8 m · s$^{$^{−2}$}$ downwards Rockets Like with lifts, rockets are also examples of objects in vertical motion. The force of gravity pulls the rocket down while the thrust of the engine pushes the rocket upwards. The force that the engine exerts must overcome the force of gravity so that the rocket can accelerate upwards. The worked example below looks at the application of Newton’s Second Law in launching a rocket. Worked Example 2: Newton II - rocket Question: A rocket is launched vertically upwards into the sky at an acceleration of 20 m·s$^{−2}$. If the mass of the rocket is 5000 kg, calculate the magnitude and direction of the thrust of the rocket?s engines. Hướng dẫn Step 1 : Analyse what is given and what is asked We have the following: m = 5000 kg a = 20 m·s$^{−2}$ F$_{g}$ = 5000 x 9,8 = 49000 N We are asked to find the thrust of the rocket engine F1. Step 2 : Find the thrust of the engine We will apply Newton’s Second Law: F$_{R}$ = ma F1 + F$_{g}$ = ma F1 + (−49000) = (5000)(20) F1 = 149 000 N upwards Worked Example 3: Rockets Question: How do rockets accelerate in space? Hướng dẫn Gas explodes inside the rocket. This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket). Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced. This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards. Worked Example 4: Newton II - lifts Question: A lift, mass 250 kg, is initially at rest on the ground floor of a tall building. Passengers with an unknown total mass, m, climb into the lift. The lift accelerates upwards at 1,6 m·s$^{−2}$. The cable supporting the lift exerts a constant upward force of 7700 N. Use g = 10 m·s$^{−2}$. Draw a labeled force diagram indicating all the forces acting on the lift while it accelerates upwards. What is the maximum mass, m, of the passengers the lift can carry in order to achieve a constant upward acceleration of 1,6 m·s$^{−2}$. Hướng dẫn Step 1 : Draw a force diagram. Step 2 : Find the mass, m. Let us look at the lift with its passengers as a unit. The mass of this unit will be (250+ m) kg and the force of the Earth pulling downwards (F$_{g}$ ) will be (250 + m) x 10. If we apply Newton’s Second Law to the situation we get: F$_{net }$= ma F$_{C}$ − F$_{g}$ = ma 7700 − (250 + m)(10) = (250 + m)(1,6) 7700 − 2500 − 10 m = 400 + 1,6 m 4800 = 11,6 m m = 413,79 kg Xem thêm: Physics 11.II Force, Momentum, Impulse High School Students Studying the Sciences Physics