Tag: force

Questions Related to force

Force acting on a particle is 5 N. If the unit of length and time are doubled and unit of mass is halved then the numerical value of the force in the new unit will be ________ times the initial value.

physics force concept of force and its unit force - push or pull forces

  1. 1

  2. 2

  3. 3

  4. 4

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Correct Option: D
Explanation:

Force = $\displaystyle 5: : \frac{kg\times m}{sec^{2}}$
If unit of length and time are doubled and the unit of mass is halved
Then the unit of force will be $\displaystyle \left ( \frac{\frac{1}{2}\times 2}{(2)^{2}} \right )=\frac{1}{4}$ times
Hence the numerical value of the force will be 4 times (as numerical value $\displaystyle \propto\frac{1}{unit}$)

1 dyne is equal to

physics force concept of force and its unit force - push or pull forces

  1. 980 g wt

  2. $\displaystyle \frac{1}{980}$ g wt

  3. 980 kg wt

  4. none of these

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Correct Option: B
Explanation:

Dyne is unit of force specified in CGS system. 

So $ 1 dyne= 10^{-5} N $
And $ 1 N = \dfrac{1}{9.80} kg $
Now,
$ 1 dyne= \dfrac{1}{9.80} \times 10^{-5}  kg$


Changing kg into grams,
 $ 1 dyne = \dfrac{1}{9.80} \times 10^{-5}  \times 10^3 g $
So,
 $ 1 dyne= \dfrac{1}{980}g  \ wt $.

Which of the following sets of concurrent forces may be in equilibrium?

physics force concept of force and its unit force - push or pull forces

  1. $\displaystyle { F } _{ 1 }=3N,{ F } _{ 2 }=5N,{ F } _{ 3 }=1N$

  2. $\displaystyle { F } _{ 1 }=3N,{ F } _{ 2 }=5N,{ F } _{ 3 }=9N$

  3. $\displaystyle { F } _{ 1 }=3N,{ F } _{ 2 }=5N,{ F } _{ 3 }=6N$

  4. $\displaystyle { F } _{ 1 }=3N,{ F } _{ 2 }=5N,{ F } _{ 3 }=15N$

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Correct Option: C
Explanation:
"Forces are in equilibrium when vector sum of these three forces is zero and they form a triangle.
For a triangle, the sum of the two sides is always greater than the third side
Therefore, $F _1 = 3N; F _2 = 5N; F _3 = 6N$

$F _1 + F _2 > F _3$

$F _2 + F _3 > F _1$

$F _3 + F _1  > F _2$


In a game of tug of war, three girls of team $A$ pull the rope with forces of $100 N$, $120 N$ and $170 N$. In team $B$, the three members pull the rope with forces of $130 N$, $150 N$ and $155 N$. Who will win the tug of war? What is the resultant force?

physics force concept of force and its unit force - push or pull forces

  1. Team $A$, $55 N$

  2. Team $B$, $55 N$

  3. Team $A$, $45 N$

  4. Team $B$, $45 N$

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Correct Option: D
Explanation:

Resultant force is nothing but the total force acting on a body including their directions. If the object is at rest or it moves with a same velocity, the resultant force should be zero. All forces are acting on same direction, resultant force should be the sum of all the forces.

In this case, the forces applied by team A are 100 N, 120 N and 170 N. The total force is 100 N+120 N+170 N = 390 N. The forces applied by tema B are 130 N, 150 N and 155 N. The total force is 130 N+150 N+155 N=435 N.
Now, both the total forces are in the opposite direction as the team are in the opposite direction. So, the resultant force is 435 - 390 = 45 N. 
Therefore, Team B will win in the game as they apply more force of 45 N.

Suppose a body that is acted on by only two forces, is accelerated. For this situation, mark the correct statement :
( Initial Velocity is non zero and in same direction as acceleration  ) 

physics force concept of force and its unit force - push or pull forces

  1. The body cannot move with constant speed

  2. The velocity can never be zero

  3. The sum of two forces cannot be zero

  4. The two forces must act in the same line

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Correct Option: A,B,C
Explanation:

If the body is accelerating, then there is a net force which drives the body. Since the body is accelerating, velocity can not be zero or moving with a constant speed. 
It is not necessary that the two forces must act in the same line however they may act in the same line.

A force F is given by $F=at+bt^2$, where t is time. The dimensions of a and b are:

physics force concept of force and its unit force - push or pull forces

  1. $[MLT^{-3}]$and$ [MLT^{-4}]$

  2. $[MLT^{-4}]$and$ [MLT^{-3}]$

  3. $[MLT^{-1}]$and$ [MLT^{-2}]$

  4. $[MLT^{-2}]$and$ [MLT^{0}]$

Show answer
Correct Option: A
Explanation:

Given that $F=at+bt^2$ where $t$ is time.


For this equation to be dimensionally correct, the each term in the right hand side should have the dimensions of $F$

Dimensional formula of $F=[ MLT^{-2}]$

$\implies aT=[ MLT^{-2}]$
$\implies$ Dimension of $a$ is $[ MLT^{-3}]$

and
$\implies bT^2=[ MLT^{-2}]$
$\implies$ Dimension of $b$ is $[ MLT^{-4}]$

Hence, option $(A)$ is correct.