Question

What is the answer for the first part of a? In kN

Answer 1

We are asked to determine the magnitude of the force that acts parallel to the nail. To do that we will add the torque that acts on the point of contact. First, we draw a free-body diagram of the situation:

We have decomposed the force of the nail into its horizontal and vertical components. The horizontal component does no torque since there is no distance parallel to the force to the point of contact.

Now, we add the torques. We consider counterclockwise as positive:

`\Sigma T=-(30cm)F+(5cm)(R_y)`

Since we consider the moment before there is no angular acceleration the sum of torques adds up to zero:

`-(30cm)F+(5cm)(R_y)=0`

Now, we determine the value of Ry as a function of "R" using the trigonometric function cosine:

`\cos \theta=(R_y)/(R)`

Now, we multiply both sides by "R":

`R\cos \theta=R_y`

Now, we substitute in the sum of torques:

`-(30cm)F+(5cm)(R\cos \theta_{})=0`

Now, we solve for "R". First, we add "30F" to both sides:

`(5cm)(R\cos \theta)=(30cm)F`

Now, we divide both sides by 5cm:

`(R\cos \theta)=((30cm)F)/(5cm)`

Now, we divide both sides by cosine:

`R=((30cm)F)/(5cm\cos \theta)`

Now, we substitute the values:

`R=((30cm)(155N))/((5cm)\cos 26.9)`

Solving the operations:

`R=1042.8N`

Therefore, the force on the nail is 1042.8 Newtons.