A Nail on a Vertical wall is Pulled by means of a Cord Attached to its head. if the Cord makes an angle of 60deg to the Horizontal and it Exerts a force of 50N on the Nail, find
a. The effective force which tends to Pull the Nail out of the wall. With Diagram.
b. The force which tends to bend the Nail
you are exerting
50 cos 60 horizontal (pulling)
and
50 sin 60 vertical (bending)
a. The effective force which tends to pull the nail out of the wall can be found by resolving the force along the vertical direction. Let's assume that the force pulling the nail out of the wall is F_out.
Using trigonometry, we can find the vertical component of the force exerted by the cord:
F_vertical = F * sin(60°)
where F is the force exerted by the cord (50N).
Therefore, the effective force which tends to pull the nail out of the wall is F_out = F_vertical = 50N * sin(60°).
b. The force that tends to bend the nail can be found by resolving the force along the horizontal direction. Let's assume that the force bending the nail is F_bend.
Using trigonometry, we can find the horizontal component of the force exerted by the cord:
F_horizontal = F * cos(60°)
where F is the force exerted by the cord (50N).
Therefore, the force that tends to bend the nail is F_bend = F_horizontal = 50N * cos(60°).
(Note: The actual values for F_out and F_bend can be calculated by substituting the numerical values into the formulas.)
a. To find the effective force that tends to pull the nail out of the wall, we need to resolve the force exerted by the cord into horizontal and vertical components.
Let's consider the following diagram:
```
F (50 N)
/|
/ |
/ |
/ |
/ θ |
/ |
/_____|______ wall
FV
```
Here, F is the force exerted by the cord (50 N), θ is the angle between the cord and the horizontal (60 degrees), and Fv is the vertical component of force.
To find Fv, we can use trigonometry. Since the angle θ is given, we can use the sine function:
sin(θ) = Fv / F
Rearranging the formula, we get:
Fv = F * sin(θ)
Fv = 50 N * sin(60 degrees)
Fv ≈ 50 N * 0.866
Fv ≈ 43.3 N
Therefore, the effective force tending to pull the nail out of the wall is approximately 43.3 N.
b. To find the force that tends to bend the nail, we need to find the horizontal component of the force exerted by the cord. Let's call it Fh.
To find Fh, we can again use trigonometry. Since the angle θ is given, we can use the cosine function:
cos(θ) = Fh / F
Rearranging the formula, we get:
Fh = F * cos(θ)
Fh = 50 N * cos(60 degrees)
Fh ≈ 50 N * 0.5
Fh ≈ 25 N
Therefore, the force that tends to bend the nail is approximately 25 N.
To find the answer to these questions, we need to resolve the forces acting on the nail and analyze the different components.
a. The effective force which tends to pull the nail out of the wall:
Let's break down the forces acting on the nail:
1. Tension force (T): This is the force exerted by the cord attached to the nail's head. It makes an angle of 60 degrees with the horizontal.
To find the vertical component of the tension force (T⊥), we can use trigonometry. The vertical component can be calculated by multiplying the tension force (T) by the sine of the angle (60 degrees):
T⊥ = T * sin(60°)
Given that the tension force (T) is 50 N:
T⊥ = 50 N * sin(60°) ≈ 43.3 N
So, the vertical component of the tension force exerted on the nail is approximately 43.3 N. This force tends to pull the nail out of the wall.
b. The force which tends to bend the nail:
To find the force that tends to bend the nail, we need to consider the horizontal component of the tension force (T⊥). This force acts perpendicular to the wall and tends to bend the nail.
Using trigonometry, we can calculate the horizontal component of the tension force (T⊃):
T⊃ = T * cos(60°)
T⊃ = 50 N * cos(60°) ≈ 25 N
So, the force that tends to bend the nail is approximately 25 N.
In summary:
a. The effective force which tends to pull the nail out of the wall is approximately 43.3 N.
b. The force that tends to bend the nail is approximately 25 N.
Please note that these calculations are based on the assumption that the nail is frictionless and the rope is massless, which allows us to neglect any additional forces such as friction or weight.