Question

A worker applies a torque to a nut with a wrench 0.400 m long. Because of the cramped space, she must exert a force upward at an angle of 52.0° with respect to a line from the nut through the end of the wrench. If the force she exerts has magnitude 34.0 N, what magnitude torque (in N · m) does she apply to the nut? (Assume that th

Answer 1

Answer:

• 10.717  N  m

Explanation:

We can find this easily if we remember the formula for the magnitude of the cross product:

$| \vec{A} \times \vec{B} | = | \vec{A} | | \vec{B} | sin (\theta)$

where $\theta$ is the angle between the vectors.

As the torque $\vec{\tau}$ is defined as:

$\vec{\tau} = \vec{r} \times \vec{F}$

where $\vec{r}$ is the position where the force $\vec{F}$ is applied. We find for our problem, that the magnitude of the torque will be:

$| \vec{\tau} | = | \vec{r} | | \vec{F} | sin (52.0 \ \°)$

$| \vec{\tau} | = 0.400 \ m *  34.0 \ N \ sin (52.0 \ \°)$

$| \vec{\tau} | = 10.717 \ N \ m$

and this is the magnitude of the torque.

Answer 2

The magnitude of a torque ls the product of the applied force and the distance turned. The magnitude of the torque applied to the nut is 10.717 Nm

Given the Parameters :

• Radius, r = 0.4 meters long

• Applied force, F = 34.0 N

• Angle turned, θ = 52°

The magnitude of the torque, τ can be defined using the relation :

• τ = rF(Sinθ)

τ = (0.4 × 34 × sin(52))

τ = 0.4 × 34 × 0.7880107

τ = 10.717 Nm

Therefore, the magnitude of the torque is 10.717 Nm

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