A block of base 10 cm times 10 cm and heigh | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The condition for sliding is $	heta > \phi$,where $\phi$ is the angle of repose. Given $\mu = \sqrt{3}$,the angle of repose is $\phi = 	an^{-1}(

Vedclass · Accepted Answer

the block will remain at rest on the plane up to certain $	heta$ and then it will topple

Vedclass · Answer

(B) The condition for sliding is $	heta > \phi$,where $\phi$ is the angle of repose. Given $\mu = \sqrt{3}$,the angle of repose is $\phi = 	an^{-1}(\mu) = 	an^{-1}(\sqrt{3}) = 60^{\circ}$.The condition for toppling is that the line of action of the weight must pass outside the base of the block. This occurs when $	an 	heta > \frac{b}{h}$,where $b$ is the base width and $h$ is the height.For the given block,$b = 10 \ cm$ and $h = 15 \ cm$. The angle at which toppling begins is $	heta_{topple} = 	an^{-1}(\frac{b}{h}) = 	an^{-1}(\frac{10}{15}) = 	an^{-1}(\frac{2}{3}) \approx 33.7^{\circ}$.Since $	heta_{topple} \approx 33.7^{\circ}$ is less than the angle of repose $\phi = 60^{\circ}$,the block will start toppling before it can start sliding. Therefore,as $	heta$ increases from $0^{\circ}$,the block remains at rest until $	heta \approx 33.7^{\circ}$,at which point it will topple.

$A$ block of base $10 \ cm \times 10 \ cm$ and height $15 \ cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then

Similar Questions

$A$ given object takes $n$ times the time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down an identical perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is:

If the coefficient of friction between an ant and the surface is $1/3$,find the maximum value of $\alpha$.

$A$ brick of mass $2 \, kg$ begins to slide down on a plane inclined at an angle of $45^\circ$ with the horizontal. The force of friction will be

$A$ body of mass $2 \ kg$ is at rest at the bottom of an inclined plane of length $8 \ m$ and height $1 \ m$ as shown in the figure. If the coefficient of friction is $0.2$,what is the work done in moving the body from the bottom to the top of the incline (in $J$)? (Take $g = 10 \ m/s^2$)

$A$ body of mass $100\, g$ is sliding down an inclined plane of inclination $30^\circ$. What is the frictional force experienced if the coefficient of friction $\mu = 1.7$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module