Two equal and opposite forces each F act on a | vedclass

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

Question

(A) Let the cross-sectional area of the rod be $a$. The area of the inclined section $AB$ is $A' = \frac{a}{\sin 	heta}$.The force $F$ acting on the

Vedclass · Accepted Answer

$\frac{F \sin 	heta \cos 	heta}{a}$

Vedclass · Answer

(A) Let the cross-sectional area of the rod be $a$. The area of the inclined section $AB$ is $A' = \frac{a}{\sin 	heta}$.The force $F$ acting on the rod can be resolved into two components along the section $AB$:$1$. Normal component: $F_n = F \sin 	heta$$2$. Tangential (shearing) component: $F_s = F \cos 	heta$The shearing stress $	au$ is defined as the tangential force divided by the area of the section:$	au = \frac{F_s}{A'} = \frac{F \cos 	heta}{a / \sin 	heta} = \frac{F \sin 	heta \cos 	heta}{a}$

Two equal and opposite forces each $F$ act on a rod of uniform cross-sectional area $a$ as shown in the figure. The shearing stress on the section $AB$ will be

Similar Questions

The breaking stress for a wire of radius $r$ of a given material is $F \ Nm^{-2}$. The breaking stress for a wire of the same material of radius $2r$ will be:

Hooke's law defines:

$A$ wire can be broken by applying a load of $20 \, kg$ weight. The force required to break the wire of twice the diameter is .......... $kg \, wt$.

The load versus elongation graph for four wires of same length and the same material is shown in the figure. The thinnest wire is represented by the line:

Two wires $A$ and $B$ are stretched by the same load. The radius of wire $A$ is double the radius of wire $B$. The stress on the wire $B$ as compared to the stress on the wire $A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module