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Physics Case study (4 Marks)

Laws of Motion · Uttarakhand Board (UBSE) STD 11 Science (Uttarakhand - English Medium). 8 questions.

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Case study (4 Marks)

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8 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Read the passage given below and answer the following questions from $1$ to $5.$ Friction: Let us return to the example of a body of mass m at rest on a horizontal table. The force of gravity $(mg)$ is cancelled by the normal reaction force $(N)$ of the table. Now suppose a force F is applied horizontally to the body. We know from experience that a small applied force may not be enough to move the body. But if the applied force F were the only external force on the body, it must move with acceleration F/m, however small. Clearly, the body remains at rest because some other force comes into play in the horizontal direction and opposes the applied force F, resulting in zero net force on the body. This force fs parallel to the surface of the body in contact with the table is known as frictional force, or simply friction. When there is no applied force, there is no static friction. It comes into play the moment there is an applied force. As the applied force $F$ increases, fs also increases, remaining equal and opposite to the applied force $($up to a certain limit$),$ keeping the body at rest. Hence, it is called static friction. Static friction opposes impending motion. The term impending motion means motion that would take place $($but does not actually take place$)$ under the applied force, if friction were absent. It is found experimentally that the limiting value of static friction $(fs )$ max f is independent of the area of contact and varies with the normal force$(N)$ approximately as: $(\text{f}_{\text{s}})\text{max}=\mu\text{N}$ where μs is a constant of proportionality depending only on the nature of the surfaces in contact. The constant μs is called the coefficient of static friction. The law of static friction may thus be written as $(\text{f}_{\text{s}})\leq\mu\text{sN}$ Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by $fk$. Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction: $(\text{f}_{\text{k}})=\mu_{\text{k}}\text{N}$
  1. Force of static friction is directly proportional to:
  1. Normal reaction
  2. Force by gravity
  3. Velocity of body
  4. None of these
  1. Coefficient of kinetic friction is independent of area of contact. True or false?
  1. True
  2. False
  1. Give formula for law of static friction
  1. Explain law of static friction
  1. Explain kinetic friction.
Answer
  1. (a) Normal reaction
  1. (a) True
  1. The law of static friction can be written as
$(\text{f}_{\text{s}})\leq\mu\text{sN}$
Where \mu s is coefficient of static friction and N is normal reaction.
  1. It is found experimentally that the limiting value of static friction $(fs)$ max $f$ is independent of the area of contact and varies with the normal force(N) approximately as: $(\text{f}_{\text{s}})\text{max}=\mu\text{N}.$ where $\mu$ s is a constant of proportionality depending only on the nature of the surfaces in contact. The constant $\mu$ s is called the coefficient of static friction. The law of static friction may thus be written as $(\text{f}_{\text{s}})\leq\mu\text{sN}$
  1. Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by $f_k$ and given by $(\text{f}_{\text{k}})=\mu_{\text{k}}\text{N}$
Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity.
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Question 24 Marks
Read the passage given below and answer the following questions from (i) to (v). This principle is a consequence of Newton’s second and third laws of motion. In an isolated system (i.e. a system having no external force), mutual forces (called internal forces) between pairs of particles in the system causes momentum change in individual particles. Let a bomb be at rest, then its momentum will be zero. If the bomb explodes into two equal parts, then the parts fly off in exactly opposite directions with same speed, so that the total momentum is still zero. Here, no external force is applied on the system of particles (bomb).
  1. A bullet of mass 10g is fired from a gun of mass 1kg with recoil velocity of gun 5m/ s. The muzzle velocity will be:
  1. 30km/ min
  2. 60km/ min
  3. 30m/ s
  4. 500m/ s
  1. A shell of mass 10 kg is moving with a velocity of 10 ms" 1 when it blasts and forms two parts of mass 9 kg and 1 kg respectively. If the first mass is stationary, the velocity of the second is:
  1. $1ms^{-1}$
  2. $10ms^{-1}$
  3. $100ms^{-1}$
  4. $1000ms^{-1}$
  1. A bullet of mass 0.1kg is fired with a speed of 100ms’ 1 . The mass of gun being 50kg, then the velocity of recoil becomes:
  1. $0.05ms^{-1}$
  2. $0.5ms^{-1}$
  3. $0.lms^{-1}$
  4. $0.2ms^{-1}$
  1. A unidirectional force F varying with time Tas shown in the figure acts on a body initially at rest for a short duration:
2.T. Then, the velocity acquired by the body is
  1. $\frac{\pi\text{F}_\text{o}\text{T}}{4\text{m}}$
  2. $\frac{\pi\text{F}_\text{o}\text{T}}{2\text{m}}$
  3. $\frac{\text{F}_\text{o}\text{T}}{4\text{m}}$
  4. $\text{zero}$
  1. Two masses ofM and 4Af are moving with equal kinetic energy. The ratio of their linear momenta is
  1. $1 : 8$
  2. $1 : 4$
  3. $1 : 2$
  4. $4 : 1$
Answer
  1. (d) 500m/ s
Explanation:
Conservation of linear momentum give
$\text{m}_1\upsilon_1+\text{m}_2\upsilon_2=0$
$\text{m}_1\upsilon_1+\text{m}_2\upsilon_2$
$\Rightarrow\upsilon_1=\frac{-\text{m}_2\upsilon_2}{\text{m}_1}$
Given,
$\text{m}_1=10\text{g}=\Big(\frac{10}{1000}\Big)\text{kg}$
$\text{m}_2=1\text{kg}\text{ and }\upsilon=-5\text{m/s}$
$\therefore$ Velocity of muzzle,
$\upsilon=\frac{+1\times5}{\frac{10}{1000}}=5\text{m/s}$
  1. (c) $100ms^{-1}$
Explanation:
Given that,
$\upsilon=10\text{m/s},$
$\text{m}_1=10\text{kg},\upsilon_2=0$
$\text{m}_2=9\text{kg},\upsilon_3=\upsilon,$
$\text{m}_3=1\text{kg}$
According to conservation of momentum,
$\text{m}_1\upsilon_1=\text{m}_2\upsilon_2+\text{m}_3\upsilon_3$
$10\times10=9\times0+1\times\upsilon$
$\Rightarrow\upsilon=100\text{ms}{^-1}$
  1. (d) $0.2ms^{-1}$
Explanation:
From the law of conservation of momentum, Initial momentum = Final momentum
$\Rightarrow\text{m}_1\upsilon_1+\text{m}_2\upsilon_2=\text{m}_1\upsilon_1+\text{m}_2\upsilon_2$
$\therefore0.1\times0+50\times0$
$=0.1\times100+50(-\upsilon_2)$
$\Rightarrow0=10-50\upsilon_2$
$\therefore\upsilon_2=\frac{10}{50}=0.2\text{ms}{^-1}$
  1. (d) $\text{zero}$
Explanation:
From 0 to T, area is positive and from T to 2T, area is negative, so net area is zero. Hence, there is no change in momentum.
  1. (c) 1 : 2
Explanation:
Two masses are moving with equal kinetic energy
$\frac{1}{2}\text{M}\upsilon^2_1=\frac{1}{2}4\text{M}\upsilon^2_2$
$\frac{\upsilon_1}{\upsilon_2}=2$
The ratio of linear momentum is
$\frac{\text{p}_1}{\text{p}_2}=\frac{\text{M}\upsilon_1}{4\text{M}\upsilon_2}$
$\frac{\text{p}_1}{\text{p}_2}\frac{1}{4}\Big(\frac{\upsilon_1}{\upsilon_2}\Big)$
$\frac{\text{p}_1}{\text{p}_2}=\frac{2}{4}=\frac{1}{2}$
$\Rightarrow\text{p}_1:\text{P}_2=1:2$
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Question 34 Marks
Read the passage given below and answer the following questions from 1 to 5. The product of force and time which is the change in momentum of the body remains a measurable quantity. This product is called impulse Impulse = Force × time duration = Change in momentum Large force acting for a short time to produce a finite change in momentum is called an impulsive force. The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. The two opposing forces are also known as action and reaction forces. Answer the following questions. The second and third laws of motion lead to an important consequence: the law of conservation of momentum. Take a familiar example. A bullet is fired from a gun. If the force on the bullet by the gun is F, the force on the gun by the bullet is – F, according to the third law. The two forces act for a common interval of time $\triangle\text{t}$ According to the second law,$\text{F}\triangle\text{t}$ is the change in momentum of the bullet and $-\text{F}\triangle\text{t}$ is the change in momentum of the gun. Since initially, both are at rest, the change in momentum equals the final momentum for each. Thus if pb is the momentum of the bullet after firing and pg is the recoil momentum of the gun, pg = – pb i.e. pb + pg = 0 That is, the total momentum of the (bullet + gun) system is conserved. Thus in an isolated system (i.e. a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged. This fact is known as the law of conservation of momentum. The total momentum of an isolated system of interacting particles is conserved.
  1. Action reaction forces acts on bodies in order that:
  1. Action acts first then reaction force comes.
  2. Reaction acts first then action force comes.
  3. Both action reaction act at same time.
  4. None of the above.
  1. Which of the following is correct about action reaction forces?
  1. They act on different objects.
  2. They are equal in magnitude and opposite in direction.
  3. Both forces acted on different object simultaneously
  4. All the above.
  1. State Newton’s third law of motion.
  1. Define impulse. Give its formula.
  1. State law of conservation of momentum.
Answer
  1. (b) Reaction acts first then action force comes.
  1. (d) All the above.
  1. The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object.
  1. The product of force and time which is the change in momentum of the body remains a measurable quantity. This product is called impulse
Impulse = Force × time duration
= Change in momentum
Large force acting for a short time to produce a finite change in momentum is called an impulsive force.
  1. The total momentum of isolated system particles is conserved. Isolated system means no external force.
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Question 44 Marks
Read the passage given below and answer the following questions from 1 to 5. Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted By p: p = m v Momentum is clearly a vector quantity. SI unit is kg m/s. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) is parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.
  • If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.
  • Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. Taken together, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied.
  1. SI unit of momentum is:
  1. Kgm/s
  2. Kgm/s2
  3. m/s2
  4. None of these
  1. Momentum is:
  1. Scalar quantity
  2. Vector quantity
  1. Define momentum. Give its SI unit.
  1. Explain with example how mass of body is important for determining effect of force on its motion?
  1. Explain with example how speed is important for determining effect of force on its motion?
Answer
  1. (a) Kgm/s
  1. (b) Vector quantity
  1. Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted By p:
P = m v
Momentum is clearly a vector quantity. SI unit is kg m/s.
  1. If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.
  1. Speed is important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time.
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Question 54 Marks
Read the passage given below and answer the following questions from 1 to 5. The first law refers to the simple case when the net external force on a body is zero. The second law of motion refers to the general situation when there is net external force acting on the body. It relates the net external force to the acceleration of the body. These qualitative observations lead to the second law of motion expressed by Newton as follow: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Thus, if under the action of a force F for time interval $\triangle\text{t},$ the velocity of a body of mass m changes from v to $\text{v}+\triangle\text{v},$ i.e. its initial momentum p = m v changes by $\triangle\text{p}=\text{m}\triangle\text{v}.$ According to the Second Law $\text{F}\alpha\frac{\triangle\text{p}}{\triangle\text{t}}$ or $\text{F}=\text{k}\frac{\triangle\text{P}}{\triangle\text{t}}$ Where k is a constant of proportionality. Mathematically, F = ma, the unit of force is $kg-m/s^2$ or Newton, which has the symbol N. Let us note at this stage some important points about the second law:
  • In the second law, F = 0 implies a = 0. The second law is obviously consistent with the first law.
  • The second law of motion is a vector law.
  • The second law of motion given by is applicable to a single point particle as well as to the rigid body but internal forces is not considered in F.
  • The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant.
  1. SI unit of force is:
  1. Newton
  2. Pascal
  3. m/s
  4. None of the above
  1. According to second law of motion The rate of change of momentum of a body is directly proportional to
  1. Velocity of body
  2. Applied force
  3. Only mass of body
  4. None of the above.
  1. The second law of motion is:
  1. Vector law
  2. Scalar law
  1. State second law of motion.
  1. Write a note on $2^{nd}$ law of motion. Enlist some deductions from $2^{nd} $ law.
Answer
  1. (a) Newton
  1. (b) Applied force
  1. (a) Vector law
  1. The second law of motion is quantitative expression of force and it states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. Mathematically, F = ma, the unit of force is $kg-m/s^2$ or Newton.
  1. The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Thus, if under the action of a force F for time interval $\triangle\text{t},$ the velocity of a body of mass m changes from v to $\text{v}+\triangle\text{v},$ i.e. its initial momentum
p = m v changes by $\triangle\text{p}=\text{m}\triangle\text{v}.$ According to the Second Law
$\text{F}\alpha\frac{\triangle\text{p}}{\triangle\text{t}}$ or $\text{F}=\text{k}\frac{\triangle\text{P}}{\triangle\text{t}}$
Where k is a constant of proportionality. Mathematically,
F = ma, the unit of force is $kg-m/s^2$ or Newton, which has the symbol N. Let us note at this stage some important points about the second law:
  • In the second law, F = 0 implies a = 0. The second law is obviously consistent with the first law.
  • The second law of motion is a vector law.
  • The second law of motion given by is applicable to a single point particle as well as to the rigid body but internal forces is not considered in F.
  • The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant.
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Question 64 Marks
Read the passage given below and answer the following questions from (i) to (v). Momentum and Newton’s Second Law of Motion Momentum of a body is the quantity of motion possessed by the body. It depends on the mass of the body and the velocity with which it moves. When a bullet is fired by a gun, it can easily pierce human tissue before coming to rest resulting in casualty. The same bullet fired with moderate speed will not cause much damage. The greater the change in momentum in a given time, the greater is the force that needs to be applied. The second law of motion refers to the general situation, where there is a net external force rating on the body.
  1. A satellite in force-free space sweeps stationary interplanetary dust at a rate $\frac{\text{dM}}{\text{dt}}=\text{dv,}$ where M is the mass, v is the dt velocity of satellite and a is a constant. What is the deceleration of the satellite?
  1. $\frac{-2\text{av}^2}{\text{M}}$
  2. $\frac{-\text{av}^2}{\text{M}}$
  3. $\text{-av}^2$
  4. $\frac{\text{av}^2}{\text{M}}$
  1. A body of mass 5 kg is moving with velocity of $\text{v} = (2\hat{\text{i}} + 6\hat{\text{j}}) ms^{-1}$ at t = 0s. After time t = 2s, velocity of body is $(10\hat{\text{i}} + 6\hat{\text{j}}) ms^{-1},$ then change in momentum of body is:
  1. $40\hat{\text{i}}\text{ kg}-\text{ms}^{-1}$
  2. $20\hat{\text{i}}\text{ kg}-\text{ms}^{-1}$
  3. $30\hat{\text{i}} \text{kg}-\text{ms}^{-1}$
  4. $(50\hat{\text{i}}+30\hat{\text{j}})\text{ Kg - ms}^{-1}$
  1. A cricket ball of mass 0.25kg with speed 10m/ s collides with a bat and returns with same speed with in 0.01s. The force acted on bat is:
  1. 25N
  2. 50N
  3. 250N
  4. 500N
  1. A stationary bomb explodes into three pieces. One piece of 2 kg mass moves with a velocity of 8 ms” 1 at right angles to the other piece of mass 1 kg moving with a velocity of 12 ms -1 . If the mass of the third piece is 0.5 kg, then its velocity is:
  1. $10ms^{-1}$
  2. $20ms^{-1}$
  3. $30ms^{-1}$
  4. $AOms^{-1}$
  1. A force of 10 N acts on a body of mass 0.5kg for 0.25s starting from rest. What is its momentum now?
  1. 0.25 N/ s
  2. 2.5 N/ s
  3. 0.5 N/ s
  4. 0.75 N/ s
Answer
  1. (d) $\frac{\text{av}^2}{\text{M}}$
Explanation:
Force, $\text{F}=\frac{\text{dq}}{\text{dt}}=\upsilon\Big[\frac{\text{dM}}{\text{dt}}\Big]=\alpha\upsilon^2$
$\Rightarrow\text{a}=\frac{\text{F}}{\text{M}}=\frac{\alpha\upsilon^2}{\text{M}}$
  1. (a) $40\hat{\text{i}}\text{ kg}-\text{ms}^{-1}$
Explanation:
Given, mass, m = 5 kg
Change in velocity, $\triangle\upsilon=\upsilon_\text{f}-\upsilon_\text{i}$
$=\Big[(10-2)\hat{\text{i}}+(6-6)\hat{\text{j}}\Big]$
Change in momentum
$=\text{m}\triangle\upsilon=5\Big[8\hat{\text{i}}\Big]​​​​$
$=40\hat{\text{i}}\text{ kg}-\text{ms}^{-1}$
  1. (d) 500N
Explanation:
Momentum, $\triangle\text{p}=2\text{m}\upsilon=2\times0.25\times10=15\text{kg-m/s}$
Force, $\text{F}=\frac{\triangle\text{p}}{\triangle\text{t}}$
$=\frac{5}{0.01}=500\text{N}$
  1. (d) $AOms^{-1}$
Explanation:
Momentum of third piece,
$\text{p}=\sqrt{\text{p}{^2_\text{x}}+\text{p}^2_\text{y}}=\sqrt{(16)^2+(12)^2}$
$=20\text{ kg}-\text{m/s}$
$\upsilon=\frac{\text{p}}{\text{m}}=\frac{20}{0.5}=40\text{m/s}$
  1. (b) 0.75 N/s
Explanation:
Given, $F = 10 N, v_i = 0,$
$m = 0.5$ kg, At $= 0.25s$
Change in momentum, $Ap = p_f - p_t ...(i)$
Also,$\triangle\text{p}=\text{F}.\triangle\text{t}....(ii)$
From Eqs. (i) and (ii), we get
$FM = pf - pi$ or $10 x 0.25 = pf -mvi$
$25 = pf — 0.5 \times 0$
$\Rightarrow pf = 2.5N/s$
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Question 74 Marks
Read the passage given below and answer the following questions from (i) to (v). Force of Friction on Connected Bodies When bodies are in contact, there are mutual contact forces satisfying the third law of motion. The component of contact force normal to the surfaces in contact is called normal reaction. The component parallel to the surfaces in contact is called friction
In the above figure, 8 kg and 6 kg are hanging stationary from a rough pulley and are about to move. They are stationary due to roughness of the pulley.
  1. Which force is acting between pulley and rope?
  1. Gravitational force
  2. Tension force
  3. Frictional force
  4. Buoyant force
  1. The normal reaction acting on the system is
  1. 8g
  2. 6g
  3. 2g
  4. 4g
  1. The tension is more on side having mass of:
  1. 8kg
  2. 6kg
  3. Same on both
  4. Nothing can be said
  1. The force of friction acting on the rope is:
  1. 20N
  2. 30N
  3. 40N
  4. 50N
  1. Coefficient of friction of the pulley is
  1. $\frac{1}{6}$
  2. $\frac{1}{7}$
  3. $\frac{1}{5}$
  4. $\frac{1}{4}$
Answer
  1. (c) Frictional force
Explanation:
Frictional force acts between pulley and rope.
  1. (d) 4g
Explanation:
The reaction force is
$R = T_1 + T_2 = (8 + 6)g = 14g$
  1. (a) 8kg
Explanation:
As, tension, T = mg
$\Rightarrow\text{T}\propto\text{M}$
So, the side having 8 kg mass will have more tension.
  1. (a) 20N
Explanation:

Due to friction, tension at all points of the thread is not alike.
$T_1 - T_2 = f$
$\Rightarrow f = 8g - 6g = 2g$
$= 28N$​​​​​​​
  1. (b) $\frac{1}{7}$
Explanation:
As, UR = f = 20N
$\text{u}=\frac{20}{\text{R}}=\frac{20}{14\times10}=\frac{1}{7}$
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Question 84 Marks
Read the passage given below and answer the following questions from 1 to 5. Newton’s first law of motion states that If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body. To summaries, if the net external force is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. This property of the body is called inertia. Inertia means ‘resistance to change’. A body does not change its state of rest or uniform motion, unless an external force compels it to change that state. In other words, all objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. Consider a book at rest on a horizontal surface. It is subject to two external forces: the force due to gravity (i.e. its weight W) acting downward and the upward force on the book by the table, the normal force R. R is a self-adjusting force. This is an example of the kind of situation mentioned above. The forces are not quite known fully but the state of motion is known. We observe the book to be at rest. Therefore, we conclude from the first law that the magnitude of R equals that of W. A statement often encountered is : Since W = R, forces cancel and, therefore, the book is at rest”. This is incorrect reasoning. The correct statement is: “Since the book is observed to be at rest, the net external force on it must be zero, according to the first law. This implies that the normal force R must be equal and opposite to the weight W”.
  1. The book on table is at rest. The force of gravity here is balanced by:
  1. Force of friction.
  2. Normal reaction by table on book.
  3. Weight of table.
  4. None of these.
  1. If no external force acts on object which is at rest. it will
  1. Remain at rest.
  2. Start to move.
  3. Both a and b can possible.
  4. None of these.
  1. Define inertia.
  1. State Newton’s first law of motion.
  1. Explain why book on table remains at rest.
Answer
  1. (b) Normal reaction by table on book.
  1. (a) Remain at rest.
  1. The tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia.
  1. Newton’s first law of motion states that If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body.
  1. Consider a book at rest on a horizontal surface. It is subject to two external forces: the force due to gravity (i.e. its weight W) acting downward and the upward force on the book by the table, the normal force R. This is an example of the kind of situation mentioned above. Magnitude of R equals that of W. This implies that the normal force R must be equal and opposite to the weight W”.
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