A force is applied at an angle of 60^circ wit | vedclass

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

Question

(A) Given: Horizontal component $F_x = 40 \ N$,Angle $	heta = 60^\circ$.The horizontal component is given by $F_x = F \cos 	heta$.Substituting the v

Vedclass · Accepted Answer

$69.28$

Vedclass · Answer

(A) Given: Horizontal component $F_x = 40 \ N$,Angle $	heta = 60^\circ$.The horizontal component is given by $F_x = F \cos 	heta$.Substituting the values: $40 = F \cos 60^\circ = F 	imes (1/2)$.Therefore,the magnitude of the force $F = 40 	imes 2 = 80 \ N$.The vertical component is given by $F_y = F \sin 	heta$.Substituting the values: $F_y = 80 \sin 60^\circ = 80 	imes (\sqrt{3}/2) = 40\sqrt{3} \ N$.Using $\sqrt{3} \approx 1.732$,we get $F_y = 40 	imes 1.732 = 69.28 \ N$.

$A$ force is applied at an angle of $60^\circ$ with the horizontal. If its horizontal component is $40 \ N$,then the vertical component is calculated to be ...... $N$.

Similar Questions

Given vector $\vec{A} = 2\hat{i} + 3\hat{j}$,the angle between $\vec{A}$ and the $y$-axis is:

The $y$-component of vector $\vec{A}$ is $+3.0 \ m$. If $\vec{A}$ makes an angle of $30^{\circ}$ counterclockwise from the positive $y$-axis,the magnitude of $\vec{A}$ is (assume $\vec{A}$ is in the $x-y$ plane).

Three vectors $\overrightarrow{OP}, \overrightarrow{OQ}$ and $\overrightarrow{OR}$ each of magnitude $A$ are acting as shown in the figure. The resultant of the three vectors is $A \sqrt{x}$. The value of $x$ is:

The angles which the vector $\vec{A} = 3\hat{i} + 6\hat{j} + 2\hat{k}$ makes with the coordinate axes are

The $X$ and $Y$ components of a force $F$ acting at $30^{\circ}$ to the $x$-axis are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module