A piece of wood with mass m = 2.7 kg is held in a vise sandwiched between two wooden jaws as shown in the figure below. A blow from a hammer drives a nail which exerts a force of 460 N on the wood. If the coefficient of static friction between the wood surfaces is 0.61, what minimum normal force must each jaw of the vise exert on the wood block to hold the block in place?

F=2•F(fr)+mg =2•μ•N +mg

N=(F+mg)/2• μ

To find the minimum normal force exerted by each jaw of the vise on the wood block, we need to consider the forces acting on the block.

The forces acting on the wood block are:
1. Weight (W) - acting vertically downwards with a magnitude of mg, where m is the mass of the block (2.7 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Normal force (N) - exerted by each jaw of the vise perpendicular to the wood block.
3. Friction force (f) - opposing the motion of the wood block.

We need to determine the static friction force first. The maximum static friction force (f_max) can be calculated using the equation:

f_max = coefficient of static friction (μ) * N

Given that the coefficient of static friction (μ) is 0.61, we can calculate the maximum static friction force:

f_max = 0.61 * N

Since the block is at rest, the force of the hammer blow (460 N) must be balanced by the static friction force (f), which is equal to the maximum static friction force (f_max).

So, we have:

f = f_max
460 N = 0.61 * N

Now, we can solve for the minimum normal force (N):

0.61 * N = 460 N
N = 460 N / 0.61

N ≈ 754.1 N

Therefore, each jaw of the vise must exert a minimum normal force of approximately 754.1 N on the wood block to hold it in place.

To solve this problem, we need to consider the forces acting on the wooden block. We have the force exerted by the hammer, the normal forces exerted by the vise jaws, and the force of static friction.

1. Let's start by considering the forces acting on the wooden block in the vertical direction (perpendicular to the surface of the vise).
- There are two normal forces, one exerted by each jaw of the vise. We'll denote the magnitude of each normal force as N.
- The weight of the wooden block acts downward with a force of mg, where m is the mass of the block (2.7 kg) and g is the acceleration due to gravity (9.8 m/s^2).

2. According to Newton's second law, the sum of the forces in the vertical direction must be zero, since the block is not accelerating vertically.
- So, we can write the equation: N + N - mg = 0
- Simplifying, we get: 2N - mg = 0

3. Now, let's consider the forces acting on the wooden block in the horizontal direction (parallel to the surface of the vise).
- The force of static friction opposes the force exerted by the hammer and prevents the block from sliding. We'll denote the magnitude of static friction force as f_s.
- The force of static friction can be calculated using the equation: f_s = μ_s * N, where μ_s is the coefficient of static friction (0.61) and N is the normal force.

4. According to Newton's second law, the sum of the forces in the horizontal direction must be zero, since the block is not accelerating horizontally.
- So, we can write the equation: f_s - F = 0, where F is the force exerted by the hammer (460 N).
- Substituting f_s = μ_s * N and simplifying, we get: μ_s * N - F = 0

Now, we have two equations:
2N - mg = 0 (Equation 1)
μ_s * N - F = 0 (Equation 2)

5. To solve for N, we can substitute mg = 2.7 kg * 9.8 m/s^2 into Equation 1:
2N - (2.7 kg * 9.8 m/s^2) = 0
2N - 26.46 N = 0
2N = 26.46 N
N = 13.23 N

Therefore, each jaw of the vise must exert a minimum normal force of 13.23 N on the wooden block to hold it in place.