Question

a Bob of mass 100 gram is suspended from the ceiling of trolley 900 gram if the trolley accelerates horizontally then string makes an angle theta is equal to 30 degree with the vertical value of force needed to provide this acceleration to trolley will be (1) 5.7N (2) 2.85N (3)5.13N (4)3.13N​

Answer 1

Answer:

(1) 5.7 N

Explanation:

1) Considering the situation with respect to lift :

Since, bob is in equilibrium with respect to lift.

That is ,

$T cos(\theta)=mg\\ \\ Tsin(\theta)=ma$

2) Dividing above two equation we got,

$tan(\theta)=\frac{a}{g}\\ \\=>a=10*tan(30^{\circ})=10*\frac{1}{\sqrt{3}} =5.7 m/s^2$

3) Now, Force needed to get this acceleration is

$F=(m_{bob}+m_{trol})*a\\=(100+900)*10^{-3}*5.7=5.7 N$

Hence, Required answer is (1) 5.7 N

Answer 2

The net force needed to accelerate the trolley is 5.7 Newton  

Explanation:

Given as :

The mass of ceiling of trolley = M = 900 gram

The mass of Bob = m = 100 gram

The trolley accelerate horizontally .

Angle made by trolley = Ф = 30°

Let The net force needed to accelerate the trolley = F  N

Let The value of g = 10 m/s

According to question

From figure

T cosФ = mg             ........1

T sinФ  = m a           .........2

Now, Dividing eq 2 by eq 1 , we get

$\dfrac{Tsin\Theta }{Tcos\Theta }$  = $\dfrac{ma}{mg}$

Or, TanФ  = $\dfrac{10}{a}$

Or, Tan 30° = $\dfrac{a}{10}$

Or, $\dfrac{1}{\sqrt{3} }$   = $\dfrac{a}{10}$

by cross multiplication

a = $\dfrac{10}{\sqrt{3} }$

Or, Acceleration = a = 5.7  m/sec²

Again

Net force needed to accelerate the trolley = Total mass × acceleration

Or, F = ( M + m ) × 5.7  m/sec²

Or,  F = ( 900 g + 100 g )  ×  5.7 m/sec²

∴     F = 1000 g × 5.7 m/sec²

i.e   F = 1 kg × 5.7 m/sec²

Or, Force = F = 5.7 Newton

So, The net force needed to accelerate the trolley = F = 5.7 N

Hence, The net force needed to accelerate the trolley is 5.7 Newton  Answer