Force Equilibrium - MCAT Chemical and Physical Foundations of Biological Systems
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A potted plant is hanging from a rope attached to a hook. The plant exerts 25N of force in the downward direction. Assume the rope is weightless.
What is the tension of the rope?
A potted plant is hanging from a rope attached to a hook. The plant exerts 25N of force in the downward direction. Assume the rope is weightless.
What is the tension of the rope?
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In order for the potted plant to stay still, the net force on the plant must be 0N. The tension of the rope is the force that allows an object to hang without falling. In order to keep the plant stationary, the tension must be equal to 25N in the upward direction.
In order for the potted plant to stay still, the net force on the plant must be 0N. The tension of the rope is the force that allows an object to hang without falling. In order to keep the plant stationary, the tension must be equal to 25N in the upward direction.
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As a skydiver jumps out of a plane, he comes to terminal velocity as air resistance brings his fall to a constant speed. At what point is the net force on the skydiver the greatest?
As a skydiver jumps out of a plane, he comes to terminal velocity as air resistance brings his fall to a constant speed. At what point is the net force on the skydiver the greatest?
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The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.
As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.
The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.
As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.
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A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the tension in the rope if the system is at equilibrium?
A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the tension in the rope if the system is at equilibrium?
Tap to reveal answer
Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.
If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.
T = (2 kg)(9.8 m/s2) = 19.6N
Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.
If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.
T = (2 kg)(9.8 m/s2) = 19.6N
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A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
Assuming the system is motionless, how many weights must be on the suspended platform?
A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
Assuming the system is motionless, how many weights must be on the suspended platform?
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Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.
Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.
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Two positively charged particles are placed near one another in an otherwise isolated environment. Their charges are 
and 
, and their masses are 
and 
. The force of gravity between these particles is negligible compared to the elecrostatic force which repels them.
At a particular moment, the magnitude of particle A's acceleration is 
. At this moment, what is the magnitude of particle B's acceleration?
Two positively charged particles are placed near one another in an otherwise isolated environment. Their charges are
At a particular moment, the magnitude of particle A's acceleration is
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Use Newton's second law, 
, to relate the forces and accelerations of the particles. Writing this out for the two charges gives the following equations.
The two charges exert equal-magnitude and opposite-direction forces on one another according to Newton's third law, so 
, or using the right side of the second law equations, 
.
Plugging in our given values allows us to solve for 
.
Use Newton's second law,
The two charges exert equal-magnitude and opposite-direction forces on one another according to Newton's third law, so
Plugging in our given values allows us to solve for
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A mass hanging in equilibrium is attached to a fixed surface on the ceiling by a spring. The mass is pulled down from the ceiling, then released and allowed to move in simple harmonic motion. The mass does not lose energy due to friction or air resistance. At what point does the mass have maximum velocity?
A mass hanging in equilibrium is attached to a fixed surface on the ceiling by a spring. The mass is pulled down from the ceiling, then released and allowed to move in simple harmonic motion. The mass does not lose energy due to friction or air resistance. At what point does the mass have maximum velocity?
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The velocity of the mass will be the greatest at the equilibrium position. Although the restoring force on the mass due to the spring will be the greatest at maximum displacement from the equilibrium position, this force is being applied to the mass as it moves in the opposite direction. As the mass moves towards the equilibrium position from maximum displacement, the restoring force due to the spring is applied in the same direction as the velocity of the mass. Because of this, the object is accelerating. After the mass passes through the equilibrium position, the spring begins to apply a force in the direction opposite to the motion of the mass, resulting in the deceleration of the mass. The mass experiences its maximum velocity at the equilibrium position.
The velocity of the mass will be the greatest at the equilibrium position. Although the restoring force on the mass due to the spring will be the greatest at maximum displacement from the equilibrium position, this force is being applied to the mass as it moves in the opposite direction. As the mass moves towards the equilibrium position from maximum displacement, the restoring force due to the spring is applied in the same direction as the velocity of the mass. Because of this, the object is accelerating. After the mass passes through the equilibrium position, the spring begins to apply a force in the direction opposite to the motion of the mass, resulting in the deceleration of the mass. The mass experiences its maximum velocity at the equilibrium position.
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What is the net force on a falling object at terminal velocity?
What is the net force on a falling object at terminal velocity?
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When an object is at terminal velocity, the upward force on the object from air resistance is equal to the force of gravity. These equal, but opposing, forces cancel each other out and the net force is zero. This results in a zero acceleration, allowing the object to fall at a constant (terminal) velocity.
When an object is at terminal velocity, the upward force on the object from air resistance is equal to the force of gravity. These equal, but opposing, forces cancel each other out and the net force is zero. This results in a zero acceleration, allowing the object to fall at a constant (terminal) velocity.
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A skydiver with a mass of 100kg jumps out of an airplane and reaches a terminal velocity of 
.
What is the total force acting on the skydiver at this point in his jump?
A skydiver with a mass of 100kg jumps out of an airplane and reaches a terminal velocity of
What is the total force acting on the skydiver at this point in his jump?
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The man has reached a terminal velocity, which means that he has an acceleration of 
Because 
, the total force on the skydiver is 0N. At this point, the force due to gravity in the downward direction is equal to the force of air resistance in the upward direction.
The man has reached a terminal velocity, which means that he has an acceleration of
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A box of mass 
is hung from the ceiling by a rope. Find the tension in the rope.
A box of mass
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There are only two forces acting on the box: the force of gravity and the force of tension in the rope. Since the box is not in motion, we can assume that the system is in equilibrium and the net force is equal to zero.
Rearranging, we can see that gravity and the force of tension will be equal and opposite.
Calculate the force of gravity using Newton's second law.
Note that the force of gravity is negative, since it acts in the downward direction. Use this value to solve for the force of tension.
There are only two forces acting on the box: the force of gravity and the force of tension in the rope. Since the box is not in motion, we can assume that the system is in equilibrium and the net force is equal to zero.
Rearranging, we can see that gravity and the force of tension will be equal and opposite.
Calculate the force of gravity using Newton's second law.
Note that the force of gravity is negative, since it acts in the downward direction. Use this value to solve for the force of tension.
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Which of the following scenarios describes a system that is not in equilibrium?
Which of the following scenarios describes a system that is not in equilibrium?
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The best way to think of a system in equilibrium is that it is displaying a constant velocity in both the horizontal and vertical directions. Both the skydiver and the tightrope walker have a constant velocity. The driver, however, is driving around a circular track. You must remember that velocity is a vector of magnitude and direction. Because the direction of the driver is changing, he is not driving at a constant velocity; thus, he is not in equilibrium, and is experiencing an acceleration (centripetal acceleration).
Keep in mind that the skydiver is experiencing both the downward force of gravity, and the upward force of air resistence on his deployed parachute. These forces cancel, producing a net force of zero, and allowing him to fall at a constant velocity in equilibrium.
The best way to think of a system in equilibrium is that it is displaying a constant velocity in both the horizontal and vertical directions. Both the skydiver and the tightrope walker have a constant velocity. The driver, however, is driving around a circular track. You must remember that velocity is a vector of magnitude and direction. Because the direction of the driver is changing, he is not driving at a constant velocity; thus, he is not in equilibrium, and is experiencing an acceleration (centripetal acceleration).
Keep in mind that the skydiver is experiencing both the downward force of gravity, and the upward force of air resistence on his deployed parachute. These forces cancel, producing a net force of zero, and allowing him to fall at a constant velocity in equilibrium.
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Three forces are acting on an object. One force acts on the object to the north, and another force acts on it to the west. The object is accelerating directly westward when all three forces are acting on it.
Based on this information, which of the following statements must be true?
Three forces are acting on an object. One force acts on the object to the north, and another force acts on it to the west. The object is accelerating directly westward when all three forces are acting on it.
Based on this information, which of the following statements must be true?
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Since the only acceleration on the object is to the west, we can conclude that there is no vertical acceleration on the object. As a result, the northern force must be offset by the third force. The only way to offset the northern force is to have a force pointing south, in any direction. As such, the third force could be directly south, southwest, or southeast as long as the southern component can counteract the northern force.
The third force would only need to be equal to the northern force if it was pointing directly due south, and we can only confirm that the third force is not completely offsetting the western force. As a result, we can only conclude that the third force has some magnitude in the southern direction.
Since the only acceleration on the object is to the west, we can conclude that there is no vertical acceleration on the object. As a result, the northern force must be offset by the third force. The only way to offset the northern force is to have a force pointing south, in any direction. As such, the third force could be directly south, southwest, or southeast as long as the southern component can counteract the northern force.
The third force would only need to be equal to the northern force if it was pointing directly due south, and we can only confirm that the third force is not completely offsetting the western force. As a result, we can only conclude that the third force has some magnitude in the southern direction.
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A man stands in the middle of a boat on a steady lake. As he walks toward one end the boat will .
A man stands in the middle of a boat on a steady lake. As he walks toward one end the boat will .
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The motion of the boat (
) needs to counteract the motion of the man (
). As the man's foot pushes him forward, the boat pushes back (Newton's third law). Due to opposing forces, the boat will move backward as the man moves forward.
This scenario can also be solved by looking at momentum. Both the man and the boat start from rest, so in order for momentum to be conserved once the man gains positive velocity, the boat must gain negative velocity.
The motion of the boat (
This scenario can also be solved by looking at momentum. Both the man and the boat start from rest, so in order for momentum to be conserved once the man gains positive velocity, the boat must gain negative velocity.
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If a wooden block is decelerating across a surface, in what direction is the net force on the object?
If a wooden block is decelerating across a surface, in what direction is the net force on the object?
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When an object is decelerating, there is a net force in the opposite direction of the motion of the object. Usually, this force is friction, which always acts counter to the net velocity. Even if there is a force in the direction of the object's motion, the net force is backwards if the object is decelerating.
Mathematically, we can prove this concept using Newton's second law. The direction of the object's motion will be positive, while the opposite (backwards) direction will be negative.
If the object is decelerating, then the acceleration is negative.
From this calculation, the force must be negative, acting counter to the direction of motion.
When an object is decelerating, there is a net force in the opposite direction of the motion of the object. Usually, this force is friction, which always acts counter to the net velocity. Even if there is a force in the direction of the object's motion, the net force is backwards if the object is decelerating.
Mathematically, we can prove this concept using Newton's second law. The direction of the object's motion will be positive, while the opposite (backwards) direction will be negative.
If the object is decelerating, then the acceleration is negative.
From this calculation, the force must be negative, acting counter to the direction of motion.
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A shuttle launched into space continues to fly in the direction it was sent even after its thrusters have turned off. What law does this illustrate?
A shuttle launched into space continues to fly in the direction it was sent even after its thrusters have turned off. What law does this illustrate?
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Newton's first law indicates that an object in motion will "want" to remain in motion at constant velocity unless it is acted on by a net force. In this case, Newton's first law is the only reasonable answer since in space, it is assumed there is no frictional or gravitational forces acting on the shuttle. Newton's second law can be described using the equation 
. Note that his first law is just a special case of the second law, where 
. Snell's law describes the interaction between waves and different media. Kirchhoff's laws involve charge conservation within circuits. Pascale's principle indicates that pressure exerted within a fluid is exerted equally in all directions.
Newton's first law indicates that an object in motion will "want" to remain in motion at constant velocity unless it is acted on by a net force. In this case, Newton's first law is the only reasonable answer since in space, it is assumed there is no frictional or gravitational forces acting on the shuttle. Newton's second law can be described using the equation
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A 
force and a 
force act at the same point on a block of wood. The 
force is pushing east and the 
force is pushing north. What is the summed force vector?
A
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Since the two vectors are perpendicular, they form a right triangle with the vector sum. Therefore, we simply use the Pythagorean theorem to solve for the sum.
The two vector components' directions determine the vector sum's direction as well. Since the components are north and east, the resultant vector is roughly northeast.
Since the two vectors are perpendicular, they form a right triangle with the vector sum. Therefore, we simply use the Pythagorean theorem to solve for the sum.
The two vector components' directions determine the vector sum's direction as well. Since the components are north and east, the resultant vector is roughly northeast.
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An object mass 
is on a table with a coefficient of kinetic friction of 
. The object is tied to another object of the same mass by a string on an ideal, frictionless pulley. The object on the table is moving to the right due to the uniform force 
.
What force needs to act on the object on the table to cause it to accelerate to the right at a constant velocity of 
?
An object mass
What force needs to act on the object on the table to cause it to accelerate to the right at a constant velocity of
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Since the block is moving to the right, we know that the frictional force is acting to the left. Also acting to the left is the gravitational force on the hanging block. Since it is a constant velocity of 
, we know that acceleration is zero and the downward force on the hanging block plus the frictional force is exactly equal to the force 
. Add the two forces.
Since the block is moving to the right, we know that the frictional force is acting to the left. Also acting to the left is the gravitational force on the hanging block. Since it is a constant velocity of
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An object mass 
is on a table with a coefficient of kinetic friction of 
. The object is tied to another object of the same mass by a string on an ideal, frictionless pulley. the object on the table is moving to the right due to the uniform force 
.
What force 
needs to act on the block on the table in order to give the two blocks an acceleration of 
to the right?
An object mass
What force
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For acceleration to be zero, the force 
needs to equal the sum of the frictional force on the block and the gravitational force on the hanging block.
To then accelerate the system, an additional force is needed.
This is the additional force required. Thus the total force required is:
For acceleration to be zero, the force
To then accelerate the system, an additional force is needed.
This is the additional force required. Thus the total force required is:
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As a skydiver jumps out of a plane, he comes to terminal velocity as air resistance brings his fall to a constant speed. At what point is the net force on the skydiver the greatest?
As a skydiver jumps out of a plane, he comes to terminal velocity as air resistance brings his fall to a constant speed. At what point is the net force on the skydiver the greatest?
Tap to reveal answer
The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.
As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.
The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.
As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.
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A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the tension in the rope if the system is at equilibrium?
A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the tension in the rope if the system is at equilibrium?
Tap to reveal answer
Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.
If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.
T = (2 kg)(9.8 m/s2) = 19.6N
Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.
If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.
T = (2 kg)(9.8 m/s2) = 19.6N
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A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
Assuming the system is motionless, how many weights must be on the suspended platform?
A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
Assuming the system is motionless, how many weights must be on the suspended platform?
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Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.
Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.
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