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Chapter 3: Problem 65

A 20 -kg wagon is pulled along the level ground by a rope inclined at \(30^{\circ}\) above the horizontal. A friction force of \(30 \mathrm{~N}\) opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and \((b)\) an acceleration of \(0.40 \mathrm{~m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
The pulling force is approximately 34.64 N at constant speed and 43.87 N during acceleration.

Step by step solution

01

Understanding Forces in Constant Motion

For the wagon moving at constant speed, the net force acting on it equals zero because there is no acceleration. Therefore, the pulling force component in the direction of motion must balance the friction force. Denote the pulling force as \(F\), and resolve it into horizontal and vertical components. The horizontal component is \(F \cos(30^{\circ})\), and the opposing friction force is 30 N.
02

Equation for Constant Speed

Set up the equation for equilibrium with constant speed: \[ F \cos(30^{\circ}) = 30 \] Thus, the pulling force can be calculated as: \[ F = \frac{30}{\cos(30^{\circ})} \approx \frac{30}{0.866} \approx 34.64 \text{ N} \]
03

Understanding Forces in Accelerated Motion

For the case where the wagon is accelerating at \(0.40 \text{ m/s}^2\), use Newton's second law \(F_{net} = ma\). Here, the net force \(F_{net}\) is the difference between the horizontal component of the pulling force and the friction, equal to the mass of the wagon times its acceleration.
04

Equation for Accelerated Motion

Set up the equation:\[ F \cos(30^{\circ}) - 30 = 20 \times 0.40 \] Simplifying gives:\[ F \cos(30^{\circ}) = 30 + 8 \] Thus, \[ F \cos(30^{\circ}) = 38 \]solve for \(F\):\[ F = \frac{38}{\cos(30^{\circ})} \approx \frac{38}{0.866} \approx 43.87 \text{ N} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Friction Force
Friction force is a resistive force that acts opposite to the direction of motion. It exists whenever two surfaces come into contact. In this problem, the friction force is given as 30 N, which opposes the movement of the wagon. The presence of friction means that more pulling force is required to either maintain a constant speed or to accelerate the wagon.

Friction can be affected by different factors:
  • The nature of the surfaces in contact: Rough surfaces typically create more friction.
  • The normal force: The force pressing the two contact surfaces together.
For an object to maintain a constant motion, the pulling force must equal the friction force. Only when these forces are equal will the net force be zero, preventing any change in speed.
Inclined Forces in Motion
Inclined forces occur when a force is applied at an angle rather than directly along the horizontal or vertical. In this problem, the pulling force is applied at a 30-degree angle to the horizontal.

Whenever forces are applied at an angle, they can be decomposed into two components:
  • The horizontal component: Calculated using the cosine of the angle times the force, it affects how effectively the force moves the object.
  • The vertical component: Calculated using the sine of the angle times the force, it may contribute to or reduce the normal force.
For the wagon, the horizontal component of the pulling force (\(F \cos(30)\)) is what counteracts the 30 N friction force or contributes to its acceleration.
Basics of Acceleration Calculations
Acceleration refers to the rate at which an object's velocity changes, calculated under Newton's second law: \[ F_{net} = ma \]where
  • \(F_{net}\) is the net force applied,
  • \(m\) is the mass of the object,
  • \(a\) is the acceleration achieved.
When the wagon is accelerating at 0.40 m/s虏, the horizontal component of the pulling force must exceed the friction force, creating a net force that causes acceleration. The needed force can be calculated by reshaping the equation to account for total forces involved, giving insight into how force, mass, and acceleration interrelate.
Determining Net Force
In physics, the net force is the overall force acting on an object resulting from the vector addition of all individual forces. It determines whether and how much an object moves.

Net force is calculated using the equation: \[ F_{net} = F_{horizontal} - F_{friction} \]
where
  • \(F_{horizontal}\) is the component of applied force in the direction of motion.
  • \(F_{friction}\) is the resistive force opposing motion.
For a moving wagon:- At a constant speed, the net force is zero as the pulling force equals the friction.- For acceleration, the net force equals the product of mass and acceleration. Thus, a sufficient pulling force must be calculated that supplies the required net force to achieve acceleration.

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Most popular questions from this chapter

A \(5.0\) -kg mass hangs at the end of a cord. Find the tension in the cord if the acceleration of the mass is ( \(a\) ) \(1.5 \mathrm{~m} / \mathrm{s}^{2}\) up, (b) \(1.5 \mathrm{~m} / \mathrm{s}^{2}\) down, (c) \(9.81 \mathrm{~m} / \mathrm{s}^{2}\) down. Don't forget gravity.

A \(400-g\) block with an initial speed of \(80 \mathrm{~cm} / \mathrm{s}\) slides along a horizontal tabletop against a friction force of \(0.70 \mathrm{~N}\). \((a)\) How far will it slide before stopping? (b) What is the coefficient of friction between the block and the tabletop? (a) Take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, \(-0.70 \mathrm{~N}\). Therefore, \(\Sigma F=m a \quad\) becomes \(\quad-0.70 \mathrm{~N}=(0.400 \mathrm{~kg})(a)\) from which \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). (Notice that \(m\) is always in kilograms.) To find the distance the block slides, make use of \(_{i x}\) \(=0.80 \mathrm{~m} / \mathrm{s}, f_{x}=0\), and \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2} .\) Then \(v_{\mathrm{K}}^{2}-u_{u}^{2}=2 a x\) yields $$ x=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 a}=\frac{(0-0.64) \mathrm{m}^{2} / \mathrm{s}^{2}}{(2)\left(-1.75 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.18 \mathrm{~m} $$ (b) Because the vertical forces on the block must cancel, the upward push of the table \(F_{N}\) must equal the weight \(m g\) of the block. Then $$ \mu_{k}=\frac{\text { Friction force }}{F_{N}}=\frac{0.70 \mathrm{~N}}{(0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.178 \text { or } 0.18 $$

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G} .\) Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$

A 600-N object is to be given an acceleration of \(0.70 \mathrm{~m} / \mathrm{s}^{2}\). How large an unbalanced force must act upon it? Notice that the weight, not the mass, of the object is given. Assuming the weight was measured on the Earth, use \(F_{W}=m g\) to find $$ m=\frac{F_{W}}{g}=\frac{600 \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=61.2 \mathrm{~kg} $$ Now that we know the mass of the object (61.2 kg) and the desired acceleration \(\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)\), the force is $$ F=m a=(61.2 \mathrm{~kg})\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)=42.8 \mathrm{~N} \text { or } 43 \mathrm{~N} $$

A 600 -kg car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. ( \(a\) ) First find the car's acceleration from a constant- \(a\) equation. It is known that \(\mathrm{v}_{i x}=30 \mathrm{v}_{f x},=0\), and \(x=70 \mathrm{~m}\). Use \(v_{k}^{2}=v_{u}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i r}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in \((a)\) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(u=F_{f} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, that $$ \begin{array}{l} F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} \\ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 \end{array} $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).
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