The lug nut on the wheel of the automobile is to be removed...

The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force of $F=30 \mathrm{Nat} A$. Determine if this force is adequate, provided $14 \mathrm{N} \cdot \mathrm{m}$ of torque about the $x$ axis is initially required to turn the nut. If the 30 -N force can be applied at $A$ in any other direction, will it be possible to turn the nut?

Transcript

00:01 Hi, so in this question we are given a wheel and we are using a lug nut on this wheel and we have to remove this using this wrench.

00:10 So this is our wrench.

00:11 Now we are applying a vertical force.

00:14 So this is the vertical force of 30 newton at this point a and this point a is at a distance of this distance is given as 0 .5 meters from the nut and the you can say the perpendicular.

00:30 Distance from the horizontal is given as 0 .3 meters so we have to find out if this much force will be adequate to remove the nut so and we are given the minimum torque that is required to move the nut so that minimum torque we will write first so the minimum torque that we require to turn the nut is tau minimum it is given as 14 newtons per meter about this point and it is about x -axis and we have to we are given the in the first case we are given a vertical force so the magnitude of the force is 30 newtons and we know that the point a is at a distance so are distance of point a from nut is given as r it is visible from here this is 0 .5 meters so this is this distance which is 0 .5 meters so we know r as well now we can make a diagram here so this is our nut this is our range and this is the point a which is at this distance is 0 .5 meters now this is the horizontal and this distance is as 0 .3 meters and this is a right angle triangle now we are applying a force f which is in this vertical along the vertical so this is the direction of force this is the direction of the radius vector so the torque we will write torque about the nut due to this force will be so the torque we will get will be tau 1 so this is let's say point o where the nut is so this will be r cross f right and this can be written as r f sine theta or rather force multiplying it by the perpendicular distance between the point o and the force so this is a right angle triangle we know the hypotenuse is 0 .5 and the perpendicular is 0 .3 so the base will be this distance will be the r perpendicular and we can calculate r perpendicular as 0 .5 squared minus 0 .3 squared this will come out to be 0 .4 meters...

Question

Please give Ace some feedback