A father is pulling his daughter uphill on a sled in slushy snow. The daughter has a mass of 22.7 kg; the sled has a mass of 3.18 kg. The coefficient of kinetic friction between the hill and the sled is 0.212 and the coefficient of static friction between the hill and the sled is 0.317. How hard does the father have to pull the sled in order to keep it moving at a constant velocity? The child holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5^{o} with the horizontal.
In this problem, you are asked to relate motion (the sled moves up the hill at a constant velocity) to force (how hard the father pulls the sled). Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem.
Step 1
Your FBD is not yet finished, because mg has both x- and y- components. Continue down to step 2 when you are ready to continue.
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Step 2
In the final FBD drawn here, all forces are divided into components. The contribution each force makes in the x-direction (along the incline) is shown explicitly, as is the contribution each force makes in the y-direction. The FBD is now a visual representation of ∑F=ma in each direction.
No. In this case, you are explicitly asked for a force and so need to consider the problem from that perspective.
The free body diagram is a sketch of the forces on an object, or the causes of motion. Acceleration is the effect of those forces and therefore does not show up on the FBD. It will make an appearance in the equation.
You can select any coordinate system that you want to use. However, the math will always be easier if you pick one axis to be in the direction of acceleration. In this case, acceleration is zero and so as a secondary consideration I used the fact that with this coordinate system I would only need to divide one force vector (gravity) into components rather than three.
There are two ways to approach this question. If you think visually, exaggerate the small and large angles on your drawing. In other words, 15.5^{o} is less than half of a right angle, so draw the slope of the incline to be very small. Then you can see that mg makes a smaller angle with the –y axis than it does with the +x axis, and the smaller angle is 15.5^{o}.
If you think geometrically, use the fact that the two small angles on a right triangle add up to 90^{o}. The triangle on the first drawing made by the ground, the incline, and mg is a right triangle, so the angle between the incline (the +x axis) and mg must be (90-15.5)^{o}=74.5^{o}. It is equally fine to use the 74.5^{o} angle, just note that sin and cos are reversed (because which side is opposite and which is adjacent depends on the triangle you draw.) You will get the same answer because sin(74.5^{o})=cos(15.5^{o}).
I chose to use the first method simply because that is an easier way to determine the correct angle for most students.
The direction of friction is always along a surface to oppose sliding. In this case, the sled is sliding uphill, and so friction acts downhill along the incline. At this stage of the problem, it is enough to know that friction is present and the direction it acts. If you also recognize that it is sliding (kinetic) friction, that is great, but you will not use that information until later in the problem.
As you can see in the figure, the x- and y-components of a vector make up the sides of a right triangle. The vector itself forms the hypotenuse (h). The side of the triangle opposite the angle that you use is given by h sinθ and the side that touches the angle you use is given by h cosθ (soh cah toa)
In this case, the x-component is opposite to the 15.5^{o} angle and adjacent to the 74.5^{o}angle, and so can be given by mg sin(15.5^{o}) as shown, or by mg cos(15.5^{o}) if you chose to use that angle instead. Likewise, the y-component is adjacent to the 15.5^{o}angle and is therefore given by mg cos(15.5^{o}).
No. Cosine goes with the component that is adjacent to θ, and sine goes with the component that is opposite to θ.
Gravity acts both to pull the sled down along the incline (in the +x direction by this drawing) and to hold the sled into the incline (in the –y direction.) Only the portion of gravity acting in the direction or interest should be included in the x- or y- equation. In other words, the x-component of gravity has no effect on how hard the sled is held into the hill.
Only three significant figures were given in the text of the problem, so only three significant figures are included in the solution.
In this problem, we are treating the child and sled as a single system. Therefore, the mass of the system is the total mass of the child and sled together, or 22.7 kg + 3.18 kg = 25.9 kg.
The direction of friction is always along a surface to oppose sliding. In this case, the sled is sliding uphill, and so friction acts downhill along the incline. At this stage of the problem, it is enough to know that friction is present and the direction it acts. If you also recognize that it is sliding (kinetic) friction, that is great, but you will not use that information until later in the problem.
All objects near Earth experience a downward force of gravity equal to mass of the object x 9.8 m/s^{2}. Because the child and the sled move together and so can be treated as a single system, m in this case refers to the total mass of the child plus the sled.
There is no information in the problem to suggest that the father pulls the rope at any particular angle to the incline. The simplest solution is if the pull is directed along the surface.
Friction always opposes the sliding motion of two surfaces across each other. Because the box slides forward across the floor, the friction of the floor pulls backward on the box.
Normal force is present whenever two objects are in contact with each other. Its direction is out from the objects and perpendicular to the surfaces. In this case, the incline pushes against the sled at an angle straight out from the surface of the incline.
There are two ways to know where to put the 15.5^{o} angle on your FBD. If you think visually, exaggerate the small and large angles on your drawing. In other words, 15.5^{o} is less than half of a right angle, so draw the slope of the incline to be very small. Then you can see that mg makes a smaller angle with the –y axis than it does with the +x axis, and the smaller angle is 15.5^{o}.
If you think geometrically, use the fact that the two small angles on a right triangle add up to 90^{o}. The triangle on the first drawing made by the ground, the incline, and mg is a right triangle, so the angle between the incline (the +x axis) and mg must be (90-15.5)^{o}. It is equally fine to use the 74.5^{o} angle that mg makes with the +x axis.
Velocity does not show up in Newton’s relation between forces and motion. Knowing velocity is constant allows you to know that acceleration is zero. You will not use this information until you put numbers into the problem.
This information tells you that the child and the sled move together. Because you don’t need to know anything about the forces between the child and the sled, you can treat them together as a single system.
As you can see in the figure, the x- and y-components of a vector make up the sides of a right triangle. The vector itself forms the hypotenuse (h). The side of the triangle opposite the angle that you use is given by h sinθ and the side that touches the angle you use is given by h cosθ (soh cah toa)
In this case, the x-component is opposite to the 15.5^{o} angle and adjacent to the 74.5^{o}angle, and so can be given by mg sin(15.5^{o}) as shown, or by mg cos(15.5^{o}) if you chose to use that angle instead. Likewise, the y-component is adjacent to the 15.5^{o}angle and is therefore given by mg cos(15.5^{o}).
The key equation for any problem that relates forces and motion is Newton’s Second Law. Regardless of what quantity you are asked to find, begin with the Second Law. If additional information is needed, it will become apparent as you proceed.
The problem asks how hard the father needs to pull the sled, so no further steps are needed to solve the problem. Note that you do not need to think about whether friction is static or kinetic until you fill in for the frictional force, and the problem would be worked in exactly the same way if you were given force and asked for acceleration, or given force and acceleration and asked for the coefficient of friction.
It does not matter whether you begin with the x- or the y-equation. If you solve for x first, you will need to pause in your solution to obtain a value for n from the y-equation.
The father pulls the daughter up the hill, which is shown to the left (or negative) direction in the figure, and so P is given a negative sign in the equation.
The normal force acts only in the +y direction, as shown on the figure.
In this problem, the sled is sliding across the hill, and so the frictional force present is kinetic, or sliding, friction. Kinetic friction is an active force (its value is given by an equation independent of other forces present) and is given by fr_{k}= µ_{k}n.
In the problem statement, you are given both kinetic and static coefficients of friction. The coefficient of static friction is unnecessary information.
Gravity acts both to pull the sled down along the incline (in the +x direction by this drawing) and to hold the sled into the incline (in the –y direction.) Only the portion of gravity acting in the direction or interest should be included in the x- or y- equation. In other words, the x-component of gravity has no effect on how hard the sled is held into the hill.
The force of gravity acts downward, which has a component in the -y direction and a component in the +x direction of this coordinate system. The correct sign is given in front of the force term in the equation. Do not also give 9.81 m/s^{2} a negative sign, or you will double count the direction. Note that 9.81 was used for g rather than 9.8, because three significant figures were given in the statement of the problem.
The child and sled are being pulled at constant velocity, and so acceleration is zero.
Only three significant figures were given in the text of the problem, so only three significant figures are included in the solution.
In this problem, we are treating the child and sled as a single system. Therefore, the mass of the system is the total mass of the child and sled together, or 22.7 kg + 3.18 kg = 25.9 kg.
You can select any coordinate system that you want to use. However, the math will always be easier if you pick one axis to be in the direction of acceleration. In this case, acceleration is zero and so as a secondary consideration I used the fact that with this coordinate system I would only need to divide one force vector (gravity) into components rather than three.
There are two ways to approach this question. If you think visually, exaggerate the small and large angles on your drawing. In other words, 15.5^{o} is less than half of a right angle, so draw the slope of the incline to be very small. Then you can see that mg makes a smaller angle with the –y axis than it does with the +x axis, and the smaller angle is 15.5^{o}.
If you think geometrically, use the fact that the two small angles on a right triangle add up to 90^{o}. The triangle on the first drawing made by the ground, the incline, and mg is a right triangle, so the angle between the incline (the +x axis) and mg must be (90-15.5)^{o}=74.5^{o}. It is equally fine to use the 74.5^{o} angle, just note that sin and cos are reversed (because which side is opposite and which is adjacent depends on the triangle you draw.) You will get the same answer because sin(74.5^{o})=cos(15.5^{o}).
I chose to use the first method simply because that is an easier way to determine the correct angle for most students.
Rotating the FBD picture is fine. It makes it easier for some people to visualize the x- and y-components of the gravitational force. Leaving the FBD oriented with the line of the incline is easier for other people to visualize how the forces relate to the motion of the sled. Use whichever orientation is better for you!
The free body diagram is a stylized drawing to help you visualize the cause of motion, and to directly map the drawing into the left hand side (ΣF) of Newton’s Second Law. Acceleration is the effect of the forces, and therefore is not included on the list of forces. Newton’s Second Law does not tie directly to velocity at all. The information given in this problem (constant velocity) allows you to know that acceleration in zero.
Either name is equally useful and you will get the same answer for that force regardless of how you think of it. I used “pull” because it is more generic and allows for different ways to picture the father pulling the sled.
There is! However, air resistance depends on velocity (squared) as well as on cross sectional area, and so at low speeds (such as pulling a sled) it is much smaller than the force of gravity on the child and sled and can therefore be ignored for the accuracy required in this problem.
It is true that air puts a small buoyant force on the child and sled. However, the size of the upward buoyant force compared to the downward force of gravity is very small. (The ratio of the two is given by the ratio of the density of air to the density of the child and sled.) So to the number of significant figures included in this problem, we do not need to take buoyant force into account.
The child and the sled move as a single system (the problem states that the child holds on and does not fall of the sled.) You are not asked for any of the forces between the child and the sled, and so you do not need to treat them separately.
The free body diagram is a sketch of the forces on an object, or the causes of motion. Acceleration is the effect of those forces and therefore does not show up on the FBD. It will make an appearance in the equation.
No. Cosine goes with the component that is adjacent to θ, and sine goes with the component that is opposite to θ.
In this problem, the child and sled are on a slight hill. Therefore, gravity does not act directly into the surface. This is seen in the figure with two components of gravity—one into the hill and one along it. It is seen in the solution because the normal force of 245 N (the supporting force of the hill) is less than the weight of the child and the sled (25.9 kg x 9.81 m/s^{2}=254 N). You expect the two numbers to be close, which they are, because the hill is only at a slight incline.
In the y-direction, the velocity of the box is also constant (the box remains at rest.) Therefore, the upward normal force from the floor exactly balances the downward pull of gravity.
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Both the equation and the figure represent the fact that the father has to balance two forces in order to keep his daughter moving at a constant velocity. He needs to balance a little bit against gravity (but not too much because the incline isn’t very steep) and he also needs to balance the force of friction that acts to slow the sled. The final answer of 120 N is less than the weight of the child and sled (254 N as seen above), and so the answer should also agree with your experience that less force is required to pull something up a snowy incline than to lift it.