Laws of Motion Test

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Laws of Motion Test - 64

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Question 1
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Two persons $$P$$ and $$Q$$ of the same height are carrying a uniform beam of length $$3 m$$. If $$Q$$ is at one end, then the distance of $$P$$ from the other end such that $$P$$, $$Q$$ receive loads in the ratio $$5 : 3$$ is:

A
$$0.5\ m$$

B
$$0.6\ m$$

C
$$0.75\ m$$

D
$$1\ m$$

Solution

$$N_{P}+N_{Q}=Mg$$
also $$\dfrac{N_{P}}{N_{Q}}=\dfrac{5}{3}$$
$$ \Rightarrow N_{P}=\dfrac{5}{8}Mg; N_{Q}=\dfrac{3}{8}Mg$$

Calculating torque about the end opposite to $$Q$$
$$N_{P}d_{1}+N_{Q}(3-d_{2})=Mg\dfrac{3}{2}$$
$$\dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg \times 3=Mg\dfrac{3}{2}$$
$$\therefore \dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg\times 3=Mg\dfrac{3}{2}$$
$$\dfrac{5}{8}Mgd_{1}=(\dfrac{12-9}{8})Mg$$
$$d_{1}=\dfrac{3}{5}=0.6\: m$$
Question 2
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Directions For Questions

Figure shows a rod of length $$\ell=1\mathrm{m}$$ and mass $$\mathrm{m}=8$$ kg. lt is hinged at the end P (on wall). At the end Q, a block of mass $$\mathrm{M}=50$$ kg is suspended. At the point R, which is at a distance equal to 25 cm from end Q, a string is connected whose other end is connected in the wall. (Given: $$sin 37^{\mathrm{o}}=\displaystyle \frac{3}{5}$$ and $$\displaystyle \cos 3\gamma=\frac{4}{5}$$) $$(\mathrm{g}$$ $$=$$ $$10\mathrm{m}/\mathrm{s}^{2})$$

...view full instructions

The horizontal component $$N_{x}$$ and vertical component $$N_{y}$$ of the reaction at end P are given by:

A
$$\mathrm{N}_{\mathrm{x}}$$ $$=$$ $$1300\mathrm{N}, \mathrm{N}_{\mathrm{y}}=960\mathrm{N}$$

B
$$\mathrm{N}_{\mathrm{x}}=960\mathrm{N}, \mathrm{N}_{y}=140\mathrm{N}$$

C
$$\mathrm{N}_{\mathrm{x}}=960\mathrm{N}, \mathrm{N}_{\mathrm{y}}=960\mathrm{N}$$

D
$$\mathrm{N}_{\mathrm{x}}=1300\mathrm{N}, \mathrm{N}_{\mathrm{y}}=1300\mathrm{N}$$

Solution

For equilibrium,$$\sum F_{x}=0$$ thus
$$Tcos37^{0}-N_{x}=0$$
$$ N_{x}=\displaystyle\frac{4}{5}T$$
$$=\displaystyle\frac{4}{5}(1200)$$
$$=960N$$
$$\sum F_{y}=0$$, thus
$$N_{y}-Tsin37^{0}+500+80=0$$
or
$$N_{y}=(1200)\left( \displaystyle\frac{3}{5} \right )-500-80$$
$$=140N$$
Question 3
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Directions For Questions

Figure shows a rod of length $$\ell=1\mathrm{m}$$ and mass $$\mathrm{m}=8$$ kg. lt is hinged at the end P (on wall). At the end Q, a block of mass $$\mathrm{M}=50$$ kg is suspended. At the point R, which is at a distance equal to 25 cm from end Q, a string is connected whose other end is connected in the wall. (Given: $$sin 37^{\mathrm{o}}=\displaystyle \frac{3}{5}$$ and $$\displaystyle \cos 3\gamma=\frac{4}{5}$$) $$(\mathrm{g}$$ $$=$$ $$10\mathrm{m}/\mathrm{s}^{2})$$

...view full instructions

The tension in the string is:

A
$$600$$ N

B
$$80$$ N

C
$$1200$$ N

D
$$800$$ N

Solution

For equilibrium about P $$\sum \tau _{p}=0$$
or
$$[(500)1+80(0.5)-(Tsin37^{0})(0.75)]=0$$
(note that torque produced by $$Tcos30^{0}$$, $$N_{x}$$ and $$N_{y}$$ about the point A is equal to zero),Thus
$$500+40-T\left ( \displaystyle\frac{3}{5} \right )(0.75)=0$$
or
$$T=\displaystyle\frac{540(5)}{3(0.75)}=1200N$$
Question 4
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A uniform of rod of length $$l$$ is placed symmetrically on two walls as shown in the figure. The rod is in equilibrium. If $$N_1$$ and $$N_2$$ are the normal forces exerted by the walls on the rod, then :

A
$$N_1 > N_2$$

B
$$N_1 < N_2$$

C
$$N_1 = N_2$$

D
$$N_1$$ and $$N_2$$ would be in the vertical directions $$N_1$$ $$N_2$$

Solution

At equilibrium, $$N_1+N_2= mg cos \ \theta$$
Also $$\tau_A= 0 \implies N_1 (0) +mg \dfrac{l}{4} cos \theta - N_2 \dfrac{l}{2}$$
$$ \dfrac{mg}{2} cos \theta = N_2 $$
Thus $$N_1= \dfrac{mg}{2} cos \theta$$
Hence $$N_1=N_2$$
Question 5
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Directions For Questions

A uniform rod of mass $$M=2 kg$$ and length $$L$$ is suspended by two smooth hinges 1 and 2 as shown in figure. A force $$F=4N$$ is applied downward at a distance $$L/4$$ from hinge 2. Due to the application of force $$F$$, hinge 2 breaks. At this instant, applied force $$F$$ is also removed. The rod starts to rotate downward about hinge 1. $$(g=10m/s^2)$$

...view full instructions

The reaction at hinge 1, before hinge 2 breaks, is:

A
$$24 \ N$$

B
$$12 \ N$$

C
$$11 \ N$$

D
$$10 \ N$$

Solution

As the rod is in equilibrium,
$$\therefore \sum F_x=0$$
$$\sum F_y=0$$
$$\sum \tau=0$$
Taking torque about hinge 2, we get
$$N_1\times L=Mg\times \cfrac {L}{2}+F\times \cfrac {L}{4}$$
$$\therefore N_1=\cfrac {Mg}{2}+\cfrac {F}{2}=11 N$$
Question 6
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Due to an impulse, the change in momentum of a body is $$1.8 \: kg \: m \: s^{1}$$. If the duration of the impulse is 0.2 s, then what is the force produced in it?

A
9 N

B
8 N

C
7 N

D
6 N

Solution
Question 7
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A ladder of length l and mass m is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is $$\mu$$. The angle $$\theta$$ at which the ladder will stay in equilibrium is:

A
$$\theta=tan^{-1} (\mu)$$

B
$$\theta=tan^{-1} (2\mu)$$

C
$$\theta=tan^{-1} (\frac {\mu}{2})$$

D
none of these

Solution

$$mg=N_1$$
$$\mu N_1=N_2$$ or $$N_2=\mu mg$$
Taking moment about O, we get
$$\mu \,m\,\,g\,\,l sin\theta-mg\dfrac {l}{2} cos\theta = 0$$
$$tan\theta=\dfrac {1}{2\mu}$$ or $$\theta=tan^{-1} \left( \dfrac {1}{2\mu} \right)$$
Question 8
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Two uniform boards, tied together with the help of a string, are balanced on a surface as shown in Fig. The coefficient of static friction between boards and surface is 0.5. The minimum value of $$\theta$$, for which this type of arrangement is possible is :

A
$$30^o$$

B
$$45^o$$

C
$$37^o$$

D
It is not possible to have this type of balanced arrangement.

Solution

For the equilibrium of the entire structure, $$N_1+N_2=2Mg$$
$$Mg(\dfrac {1}{2} cos\theta)+N_2\times 2x-Mg(\dfrac {1}{2} cos\theta+2x)=0$$
Given $$N_2=Mg, N_1=Mg$$

For individual boards, $$T_1=f_1, N_1=Mg$$
and $$T_1\times l sin \theta=Mg\dfrac {1}{2} cos\theta$$
$$f_1=T_1=\dfrac {Mg}{2 tan\theta}$$

For safe equilibrium, $$f_1 < f_L=\mu_s Mg$$
$$\dfrac {Mg}{2 tan\theta}=\mu_s Mg\Rightarrow tan\theta > \dfrac {1}{2\mu_s}\Rightarrow \theta=45^o$$
So the minimum value of $$\theta$$ for this type of arrangement is $$45^o$$
Question 9
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A vehicle is moving with a velocity $$v$$ on a curved road of width $$b$$ and radius of curvature $$R$$. For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edge of the road is

A
$$\dfrac{v^2b}{Rg}$$

B
$$\dfrac{vb}{Rg}$$

C
$$\dfrac{vb^2}{Rg}$$

D
$$\dfrac{vb}{R^2g}$$

Solution

We know that for the vehicle banking
on a curved path, $$tan\theta=\dfrac{v^{2}}{rg}$$.
Here, the vehicle moves with velocity
$$v$$ and radius of curvature $$R$$ and width is b.
So, here $$\theta$$ is small (refer figure) and thus, $$sin\theta=\dfrac{h}{b}=tan\theta$$.
Now, we can equate $$\dfrac{h}{b}=\dfrac{v^{2}}{Rg}$$
$$\Rightarrow h= \dfrac{bv^{2}}{Rg}$$, will be the required elevation between the outer and inner edges of the road.
Question 10
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When a bus starts suddenly, the passengers are pushed back. This is an example of which of the following?

A
Newton's first law

B
Newton's second law

C
Newton's third law

D
None of Newton's laws

Solution