The figure below shows a claw hammer being used to pull a nail out of a horizontal board where θ = 32.2°. The mass of the hammer is 1.00 kg. A force of 125 N is exerted horizontally as shown, and the nail does not yet move relative to the board. Assume the force the hammer exerts on the nail is parallel to the nail. the contact point is 30 cm from the force and the head of the nail is 5 cm from the contact point. what is the magnitude and direction of the force exerted by the hammer claws on the nail

To find the magnitude of the force exerted by the hammer claws on the nail, we can use the principle of moments (torque) about the point of contact between the hammer and the board. The torque due to the horizontal force (125 N) is equal to the torque due to the force exerted by the hammer claws.

Let F be the force exerted by the hammer claws on the nail.

Torque due to horizontal force = Torque due to force by hammer claws
125 N * 30 cm * sin(32.2°) = F * 5 cm

Solve for F:

F = (125 N * 30 cm * sin(32.2°)) / 5 cm
F ≈ 201.83 N

The magnitude of the force exerted by the hammer claws on the nail is approximately 201.83 N.

To find the direction of the force exerted by the hammer claws, we need to find the angle between the force and the nail. This angle can be found using the arctangent function. Since the contact point is 30 cm from the force and the head of the nail is 5 cm from the contact point, we have:

tan(angle) = (5 cm) / (30 cm)
angle = arctan(5/30)

angle ≈ 9.46°

The direction of the force exerted by the hammer claws on the nail is approximately 9.46° from the nail.

To determine the magnitude and direction of the force exerted by the hammer claws on the nail, we can break down the given information and solve step-by-step.

Step 1: Find the vertical component of the force exerted by the hammer claws on the nail.
The vertical component of the force can be found using the mass of the hammer, gravitational acceleration, and the angle θ.

Force vertical (Fv) = mass x gravity x sin(θ)

Given:
Mass (m) = 1.00 kg
Gravity (g) = 9.8 m/s²
Angle (θ) = 32.2°

Converting angle from degrees to radians:
θ (in radians) = θ (in degrees) x π / 180
θ (in radians) = 32.2° x π / 180 ≈ 0.562 rad

Substituting the given values to calculate the vertical component of the force:
Fv = 1.00 kg x 9.8 m/s² x sin(0.562 rad)

Fv ≈ 5.37 N

Step 2: Find the horizontal component of the force exerted by the hammer claws on the nail.
The horizontal component of the force can be found using the mass of the hammer, gravitational acceleration, and the angle θ.

Force horizontal (Fh) = mass x gravity x cos(θ)

Using the given values:
Fh = 1.00 kg x 9.8 m/s² x cos(0.562 rad)

Fh ≈ 8.47 N

Step 3: Calculate the perpendicular distance between the contact point and the hammer's force.
Given that the contact point is 30 cm from the force and the head of the nail is 5 cm from the contact point, we can see that the perpendicular distance is the sum of these two distances.

Perpendicular distance (d) = 30 cm + 5 cm
d = 35 cm

Converting the distance to meters:
d = 35 cm x 1 m / 100 cm
d = 0.35 m

Step 4: Calculate the torque exerted by the hammer claws on the nail.
The torque (τ) can be calculated using the following formula:

τ = force x distance

τ = Fh x d

Substituting the given values:
τ = 8.47 N x 0.35 m

τ ≈ 2.96 N·m

Step 5: Determine the magnitude and direction of the force exerted by the hammer claws on the nail.
Since the nail does not move, the force exerted by the hammer claws on the nail must be equal and opposite to the force exerted by the nail on the hammer claws. Therefore, the magnitude of the force exerted by the hammer claws on the nail is 8.47 N.

To find the direction of the force, we need to consider the vertical and horizontal components. The vertical component was found to be 5.37 N, and the horizontal component was found to be 8.47 N.

Using these components, the direction of the force exerted by the hammer claws on the nail can be determined using trigonometry:

θ = arctan(Fv / Fh) = arctan(5.37 N / 8.47 N)

θ ≈ 33.7°

Therefore, the magnitude of the force exerted by the hammer claws on the nail is 8.47 N, and the direction of the force is approximately 33.7° above the horizontal direction.

To determine the magnitude and direction of the force exerted by the hammer claws on the nail, we need to resolve the horizontal force exerted by the hammer into its components.

First, let's calculate the horizontal component of the force exerted by the hammer. Given that a force of 125 N is exerted horizontally, the horizontal component of this force can be calculated using trigonometry:

Force_horizontal = Force × cos(θ)
Force_horizontal = 125 N × cos(32.2°)
Force_horizontal ≈ 105.27 N

Now, let's consider the distances involved. The contact point is 30 cm away from the force, and the head of the nail is 5 cm away from the contact point.

We can use the concept of torque to calculate the force exerted by the hammer claws on the nail. Torque is defined as the product of the force and the distance from the pivot point (or axis of rotation). In this case, the pivot point is the contact point of the hammer with the nail.

The torque equation is given by:
Torque = Force × perpendicular distance

The perpendicular distance can be determined by subtracting the distance between the contact point and the head of the nail from the distance between the contact point and the force applied. So, the perpendicular distance would be (30 cm - 5 cm).

Perpendicular distance = 30 cm - 5 cm = 25 cm = 0.25 m

To find the force exerted by the hammer claws, we can rearrange the torque equation to solve for Force:
Force = Torque / perpendicular distance

Now, substituting the known values into the equation:
Force = Force_horizontal × perpendicular distance
Force = 105.27 N × 0.25 m
Force ≈ 26.32 N

Therefore, the magnitude of the force exerted by the hammer claws on the nail is approximately 26.32 N.

As for the direction, the force will be in the opposite direction of the applied force because the hammer is attempting to pull the nail out of the board. So, the force exerted by the hammer claws on the nail is in the opposite direction of the applied force, thus making it horizontal and directed towards the left side of the diagram.