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1. A box is sliding up an incline that makes an angle of 20 degrees with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.2. The initial speed of the box at the bottom of the incline is 2 m/s. How far does the box travel along the incline before coming to rest? Solution: The first part in the problem is to find an acceleration of the motion. The acceleration is due to gravitation force and the friction force and has the following form: The second part is to write down the kinematic equations of motion. In this problem we need to use the relation between the traveled distance and initial and final (the final velocity is 0) velocities: where s is the traveled distance. Then 2. A block weighing 80 N rests on a plane inclined at 30 degrees to the horizontal. The coefficients of static and kinetic friction are 0.2 and 0.1 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping? Solution: The minimum force corresponds to the condition that the static friction force has the maximum value, which is 0.2*normal force. To find the normal force and the external force we need to write down the condition of equilibrium: the net force is 0. Then we rewrite this equation in terms of x and y-components (x axis is parallel to the plane). The x-component of the second Newton's law has the form: The y-component: Then since Then , we obtain 3. The value of g at the surface of the earth is 9.78 N/kg, and on the surface of Venus the magnitude of g is 8.6 N/kg. A cosmonaut has a mass of 60 kg on the surface of the earth. What will her weight be on the surface of Venus? Solution: The weight is the product of the mass of cosmonaut and the free fall acceleration. On the surface of Venus: 4. A car (m=2000 kg) is parked on a road that rises 20 degrees above the horizontal. What are the magnitudes of (1) the normal force and (2) the static frictional force that the ground exerts on the tires? Solution: We need to write down the second Newton's law: the net force is 0. This is the vector equation. In the present problem there are three forces acting on the car: gravitational force, normal force, and the static friction force. Then we rewrite the second Newton's law in terms of x and y-components (x axis is parallel to the road). The x-component of the second Newton's law has the form: The y-component: From these equations we can find the normal force: and the static friction force: 5. A 100 kg load is placed on a 40 kg stretcher to be carried by two persons. The stretcher is 2m long and the load is placed 0.6m from the left end. How much force must each person exert to carry the stretcher? Solution: In this problem we need to use the condition of equilibrium for stretcher, which mean there is no translational motion and there is no rotation of the stretcher. First, we need to show all forces acting on stretcher. It is important to show not only the directions of the forces but also their application points. We characterize the position of the application point in terms of the distance to point A (shown in the figure below): 1. Force due to person 1. Application point of the force is one end of the stretcher (see figure). The distance to point A is . 2. Force due to person 2. Application point of the force is another end of the stretcher (see figure). The application point is point A. 3. Gravitational force on the stretcher. Application point of the force is the center of the mass of the stretcher, which is exactly in the middle of the stretcher – point O. Since the length of the stretcher is 2 m then the distance between point O and point A is . 4. Gravitational force on the load. Application point of the force is the center of the mass of the load. The distance between the load and the point A is . The stretcher is in equilibrium (static or dynamics equilibrium). We have two equations, which describe the condition of equilibrium: (1) no translational motion – the net force is zero: We rewrite this vector equation in terms of y-components (axis y is shown in the figure): Then ............................................(1) (2) no rotation – the net torque is zero. We can choose any point to write down the condition that the net torque about this point is zero. We choose point A. Then the net torque about point A is We substitute the known values and obtain Now we can find force : And then force (from equation (1)): 6. A painter weighing 900 N stands on a massless plank 5 m long that is supported at each end by a stepladder. If he stands 1m from one end of the plank, what force is exerted by each stepladder? Solution: This problem is similar to Problem 43. The main difference is that now the plank is massless. Therefore, in the present problem we have only three forces acting on the plank. We need to use the condition of equilibrium for the plank: no translational motion and no rotation of the palnk. First, we need to show all forces acting on the plank. It is important to show not only the directions of the forces but also their application points. We characterize the position of the application point in terms of the distance to point A (shown in the figure below): 1. Force due to stepladder 1. Application point of the force is one end of the plank (see figure). The distance to point A is . 2. Force due to stepladder 2. Application point of the force is another end of the plank (see figure) – this is the point A. 3. Gravitational force on the painter. Application point of the force is the center of the mass of the painter. The distance between the painter and the point A is . The plank is in equilibrium (static or dynamics equilibrium). There are two conditions of equilibrium: (1) no translational motion – the net force is zero: We rewrite this vector equation in terms of y-components (axis y is shown in the figure): Then ..............................................................(1) (2) no rotation – the net torque is zero. We can choose any point to write down the condition that the net torque about this point is zero. We choose point A. Then the net torque about point A is zero: We substitute the known values and obtain Now we can find force : And then force (from equation (1)) is 7. A 20 kg child and a 30 kg child sit at opposite ends of a 4 m seesaw that is pivoted at its center. Where should another 20 kg child sit in order to balance the seesaw? Solution: In this problem we need to use only one equilibrium condition: no rotation of the seesaw (the seesaw is balanced). The second equilibrium condition (no translational motion) needs to be used only if we need to find the normal force at the pivot. We can introduce any point as the point of possible rotation. We consider point O – the center of the seesaw – as the point of possible rotation. There is no rotation about point O, which means that the net torque about point O is zero. First, we need to show all forces acting on the seesaw: 1. Gravitational force (more exactly it is normal force, which is equal to the gravitational force) on the 20 kg child. Application point of the force is one end of the seesaw (see figure). The distance to point O is . 2. Gravitational force (more exactly it is normal force, which is equal to the gravitational force) on the 30 kg child. Application point of the force is the other end of the seesaw (see figure). The distance to point O is . 3. We place another 20 kg child on the seesaw. We characterize the position of the child by the distance to point O (this is an unknown distance): . The corresponding force is the gravi tational force on the child. There is no rotation about point O. Then the net torque about point O is zero: We substitute the known values and obtain From this equation we can find the position of the 20 kg child: 8. A 20 meters long rope attached at the top and the bottom of a flag pole is pulled 2 meters away from the pole by a 100 newton force acting at right angles to the pole at its mid point. What is the tension on the segments of the rope on each side of the 100 newton force? Solution: In problem we need to use the equilibrium condition for point A: the vector sum of all forces is zero. There are three forces acting on point A: an external 100 N force and two tension forces . The directions of the tension forces are along the ropes. These forces are shown in the figure below. The y component of the external force is zero and the y-components of the tension forces cancel each other. Then the condition that the x-component of the sum of all these forces is zero takes the form: Then The angle can be found from the right triangle (with lengths of 10 m and 2 m), shown in the figure: Then 9. An automobile weighing 3200 lbs. is on a road which rises 10 ft. for each 100 feet of road. What force tends to move the car down the hill? Solution: All variables in the problem should be expressed in the same units (the same system of units). It is better to use the SI units. In the SI system of units, the mass should be measured in kg and the distance is in meters. Then: The force acting on the car are the following: gravitation force, normal force, friction force, and the force pulling the force up along the hill. The friction force is the rolling friction force, which is small. The normal force does not have a component along the hill (the normal force is orthogonal to the incline). Then there is only one force pulling the car down the hill. This is the gravitational force, shown in the picture below. The magnitude of the gravitation force is Then from the picture of the incline we can find the component of the gravitational force along the inline. This component can be expressed in terms of the angle of incline, : To find the angle of incline we need to use the right triangle, shown in the picture above: Then 10. A tension of 6000 Newtons is experienced by the elevator cable of an elevator moving upwards with an acceleration of Solution: From the second Newton's law we have . What is the mass of the elevator? Then