$\begin{array}{1 1}3.76\;N\\36\;N\\29.4\;N \\46\;N\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Step 2:

Apply $\sum \overrightarrow{T}=0$ about an axis through the center of the sphere :

Apply $\sum F_x=0$ to the sphere

Substitute for f and solve for T :

Substitute numerical values and evaluate T .

$fR -TR =0 => T =f$

$f+T \cos \theta- Mg \sin \theta=0$

$T= \large\frac{Mg \sin \theta}{1+\cos \theta}$

$ T= \large\frac{(3 \;kg)(9.81 \; m/s^2) \sin 30^{\circ}}{1+\cos 30^{\circ}}$$=7.89\;N$

Step 3:

Apply $\sum F_y =0$ to the sphere :

Solve for $F_n$

Substitute numerical values and evaluate $F_n$ :

$F_n-T \sin \theta - Mg \cos \theta =0$

$F_n =T \sin \theta+Mg \cos \theta$

$F_n= (7.89 \;N) \sin 30^{\circ}+ (3 \;kg)(9.81\;m/s^2) \cos 30^{\circ}$

$\qquad= 29.4 \;N$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...