A force of 32 N is applied at the end of a wrench handle that is 27 cm long. The force is applied in a direction perpendicular to the handle as in the diagram.

(a) What is the torque applied to the nut by the wrench?
(b) What would the torque be if the force were applied half way up the handle instead of at the end?

a.)N= r*F

(32N)(.27m)= 8.64 N.m

b.) half of .27=0.135
t= F*d
(32N)(0.135)= 4.32 N.m

a) Ah, torque, the wrench's way of saying, "I'm gonna twist you real good!" To calculate the torque applied to the nut, we use the formula: torque = force × distance. In this case, the force is 32 N and the distance is 27 cm (or 0.27 m, if we want to be all metric). So, torque = 32 N × 0.27 m = 8.64 Nm. That nut better brace itself!

b) If we move the force halfway up the handle, it's like saying, "Hey, wrench, share the twisty love." The new distance from the force to the nut is now half of the original 27 cm (or 0.135 m). But the force is still the same, 32 N. So, the new torque would be 32 N × 0.135 m = 4.32 Nm. Not quite as twist-worthy, but still enough to loosen that stubborn bolt.

To find the torque applied to the nut by the wrench, we can use the formula:

Torque (τ) = Force (F) × Distance (r) × sinθ

where:
- Force (F) is the force applied at the end of the wrench handle,
- Distance (r) is the length of the wrench handle, and
- θ is the angle between the direction of the force and the direction of the handle. In this case, since the force is applied perpendicular to the handle, θ = 90 degrees.

(a) What is the torque applied to the nut by the wrench?

Given:
Force (F) = 32 N
Distance (r) = 27 cm = 0.27 m
θ = 90 degrees

Plugging the values into the formula, we get:

τ = F × r × sinθ
= 32 N × 0.27 m × sin90°
= 8.64 N·m

Therefore, the torque applied to the nut by the wrench is 8.64 N·m.

(b) What would the torque be if the force were applied halfway up the handle instead of at the end?

In this case, the distance (r) would be half of the original length, which is 0.27 m / 2 = 0.135 m.

Using the same formula, we can calculate the torque:

τ = F × r × sinθ
= 32 N × 0.135 m × sin90°
= 4.32 N·m

Therefore, if the force were applied halfway up the handle instead of at the end, the torque applied to the nut by the wrench would be 4.32 N·m.

To calculate the torque applied by the wrench, you need to use the formula:

Torque = Force × Distance × Sin(θ)

Where:
- Torque is the twisting or rotational force applied by the wrench (measured in newton-meters or Nm).
- Force is the applied force on the wrench (measured in newtons or N).
- Distance is the distance between the applied force and the axis of rotation (measured in meters or m).
- θ is the angle between the direction of the force and the direction perpendicular to the handle.

(a) Torque applied to the nut by the wrench:
In this case, the force is applied at the end of the wrench handle, so the distance is the length of the handle:

Distance = 27 cm = 0.27 m

The force is applied perpendicular to the handle, so the angle is 90 degrees (π/2 radians).

Therefore, using the formula for torque:

Torque = 32 N × 0.27 m × Sin(π/2)
= 8.64 Nm

The torque applied to the nut by the wrench is 8.64 Nm.

(b) Torque with force applied halfway up the handle:
If the force is applied halfway up the handle, the distance from the axis of rotation is only half the length of the handle:

Distance = 27 cm ÷ 2 = 0.135 m

Again, the force is applied perpendicular to the handle, so the angle is still 90 degrees (π/2 radians).

Using the formula for torque:

Torque = 32 N × 0.135 m × Sin(π/2)
= 4.32 Nm

If the force were applied halfway up the handle instead of at the end, the torque applied to the nut by the wrench would be 4.32 Nm.