Q#1
A body of mass m is placed on a table. The earth is pulling the body with a force mg. Taking this force to be the action, what is the reaction?Answer:
This action force mg is being applied by the body on the table. To hold the body in position the table also exerts equal and opposite force on the body which is the reaction. So in this case, downward force mg by the body on the table is action and the upward force mg by the table on the body is the reaction.
Q#2
A boy is sitting on a chair placed on the floor of a room. Write as many action-reaction pairs of forces as you can.
Answer:
Let the weight of the boy be W. If this force W be taken as action (downward) on the chair. Equal and opposite force W (Upward) is applied to the boy by the chair which is a reaction. So, this is the first action-reaction pair W – W.
Now this weight W plus the weight of the chair W' applies force on the floor through the four legs of the chair. Let P = W + W'.
The force P is divided into four forces P1, P2, P3 and P4 (according to the sitting posture of the boy) which act downwards through the legs of the chair.
If these be taken as actions, equal and opposite (upwards) forces by the floor on the legs will be reactions. So other pairs of action-reaction are P1 – P1, P2 – P2, P3 – P3, and P4 – P4.
Q#3
A lawyer alleges in court that the police has forced his client to issue a statement of confession. What kind of force is this?
Answer:
The lawyer means that the police has threatened his client to issue a statement of confession. So this force is not a physical force but an emotional one.
Q#4
When you hold a pen and write on your notebook, what kind of force is exerted by you on the pen? By the pen on the notebook? By you on the notebook.
Answer:
When two bodies are in contact and forces of push, pull or friction come into the picture, these forces are categorized in electromagnetic forces. So the push and friction forces exerted by us on the pen, the push force exerted by the pen on the notebook are all kind of electromagnetic force. When we write, the root of the palm exerts a push on the notebook which again is a kind of electromagnetic force. If we do not consider this then we do not directly exert forces on the notebook, it is through the pen.
Q#5
Is it true that the reaction of a gravitational force is always gravitational, of an electromagnetic force is always electromagnetic and so on?
Answer: Yes.
Q#6
Suppose the magnitude of nuclear forces between two protons varies with the distance between them as shown in figure (4-Q1). Estimate the ratio "Nuclear force/Coulomb force" for (a) x = 8 fm (b) x = 4 fm, (c) x = 2fm and (d) x = 1fm (1 fm = 10–15 m).
In the given figure force on the Y-axis is plotted on a logarithmic scale. Magnitude of Nuclear forces from the given figure by logarithmic interpolations are,
For x = 8 fm, → 0.18 N
For x = 4 fm, → 1.00 N
For x = 2 fm, → 17.7 N
For x = 1 fm, → 177 N
Now Coulomb force is given by,
F = kq²/r² = 9 x 109(1.6 x 10-19)²/r²
= 144/r² N
So this force,
for x = 8 fm, → 2.25 N
for x = 4 fm, → 9.00 N
for x = 2 fm, → 36.00 N
for x = 1 fm, → 144.0 N
So the ratio "Nuclear force/Coulomb force"
(a) for x = 8 fm is 0.18 N/2.25 N = 0.08
(b) for x = 4 fm is 1.0 N/9.0 N = 0.11
(c) for x = 2 fm is 17.7 N/36.0 N = 0.491
(d) for x = 1 fm is 177 N/144 N = 1.23
It is clear that in the range of r < 2 fm Nuclear forces are quite comparable to Coulomb forces.
Q#7
List all the forces acting on the block B in figure (4-Q2).
Forces acting on Block B are shown below,
Let us take the weight of block B as W. This force of weight W acts on block B. Since the man exerts force (say F) above the level of block B there may be movement tendency between both blocks and a force of friction (P) will be applied by block A on block B downwards. There will be a force R acting upwards on block B by the floor such that R = W + P.
The push force F by the man on block A which is transferred to block B through A is counterbalanced by a force of friction between block B and the floor. It is equal but opposite in direction to F.
Q#8
List all the forces acting on (a) pulley A, (b) the boy and (c) the block C in the figure (4-Q3).
The push force F by the man on block A which is transferred to block B through A is counterbalanced by a force of friction between block B and the floor. It is equal but opposite in direction to F.
Q#8
List all the forces acting on (a) pulley A, (b) the boy and (c) the block C in the figure (4-Q3).
(a) For forces acting on pulley A, see the figure below:
Let us take the weight of block C as W. This force W is transmitted as tension in the string throughout. So, in a balanced state, the strings on both sides of the pulley A will apply forces W horizontally and W vertically downward. Taking the pulley weightless and frictionless, the hinge of the pulley will exert a force on the pulley that is equal in magnitude but opposite in direction to the resultant of two perpendicular forces W and W.
As shown in the figure it will be equal to √2W and 45° from horizontal/vertical.
(b) The forces on the boy are shown in the figure,
The string exerts a force W on the boy upwards, the weight F of the boy acts downwards and a force R is exerted by the floor on the boy upwards such that F = W+R. Since the string exerts the force away from the body (or C.G,) of the boy resulting a slipping tendency at the feet touching the floor, it is counterbalanced by a force of friction P (Horizontal) by the floor at the feet of the boy.
(c) The forces acting on the block C are shown in the figure,
Since the block C hangs with the string, the forces acting on it are weight W of the block C downwards and the counterbalancing force W by the string upwards.
Q#9
Figure (4-Q4) shows a boy pulling a wagon on a road. List as many forces as you can which are relevant to this figure. Find the pairs of forces connected by Newton's third law of motion.
As shown in the figure it will be equal to √2W and 45° from horizontal/vertical.
(b) The forces on the boy are shown in the figure,
The string exerts a force W on the boy upwards, the weight F of the boy acts downwards and a force R is exerted by the floor on the boy upwards such that F = W+R. Since the string exerts the force away from the body (or C.G,) of the boy resulting a slipping tendency at the feet touching the floor, it is counterbalanced by a force of friction P (Horizontal) by the floor at the feet of the boy.
(c) The forces acting on the block C are shown in the figure,
Since the block C hangs with the string, the forces acting on it are weight W of the block C downwards and the counterbalancing force W by the string upwards.
Q#9
Figure (4-Q4) shows a boy pulling a wagon on a road. List as many forces as you can which are relevant to this figure. Find the pairs of forces connected by Newton's third law of motion.
The various forces acting in the figure are shown figure,
Weight of wagon W is distributed on wheels as W1 and W2. R1 and R2 are the forces exerted by the floor on the wagon wheels. Forces of friction P1 and P2 are exerted by the wheels on the floor and equal and opposite forces are exerted by the floors on the wheels. Tension force T in the string is exerted by the hand of boy and equal but opposite force is exerted by the string on the hand of the boy. This force T is also exerted by the boy through string on the wagon and equal but opposite force T is exerted by the wagon on the boy. Resultant of forces in the string at the shoulder of the boy is counterbalanced by force R on the string by the shoulder of the boy. Weight W3 of the boy acts downwards while an upward force R3 is applied by the floor on the boy. The backward push on the floor by the boy results in forces of friction P3 and equal but opposite force P3 is acted by the floor on the boy.
Pairs of forces connected by Newton's third law of motion are T – T, P1 – P1, P2 – P2, P3 – P3, and R – R. Though other forces are also connected by Newton's third law of motion they do not show as pairs in the figure. Weight W1 and W2 are distributed on the wheels according to loading patterns and C.G. of the wagon and the floor reactions R1 and R2 are according to the net force downwards on the wheels.
Q#10
Figure (4-Q5) shows a cart. Complete the table shown below.
Answer:
The table is completed below according to the following figure:
Pairs of forces connected by Newton's third law of motion are T – T, P1 – P1, P2 – P2, P3 – P3, and R – R. Though other forces are also connected by Newton's third law of motion they do not show as pairs in the figure. Weight W1 and W2 are distributed on the wheels according to loading patterns and C.G. of the wagon and the floor reactions R1 and R2 are according to the net force downwards on the wheels.
Q#10
Figure (4-Q5) shows a cart. Complete the table shown below.
Force on | Force by | Name of the force | Direction |
Cart | 1. 2. 3. | ||
Horse | 1. 2. 3. | ||
Driver | 1. 2. 3. |
Answer:
The table is completed below according to the following figure:
Force on | Force by | Name of the force | Direction |
Cart | 1.Road 2.Road 3.Horse 4.Driver 5.Driver 6.Gravity | R F P W F1 W3 | Upwards Horizontally right Along the rod on the horse, left Downward Horizontally left Downwards |
Horse | 1.Cart 2.Driver 3.Gravity 4.Road 5.Road | P H W1 W2 F1 | Along the rod on the horse, right Along the rope, right downwards Upwards Horizontally left |
Driver | 1.Horse 2.Cart 3. Cart 4.Gravity | H R1 F1 W | Along the rope, Left Upwards Right Downwards |
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