A force of 5 N is applied to the end of a lever that has a length

Physics

1 answer:

Klio2033 [76]3 years ago

4 0

Answer:

The answer is 10Nm

Explanation: I ended up just messing around with the numbers, I multiplied 5 and 2 got 10 as my answer and it was right.

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When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 43.

zmey [24]

To solve this problem, we must take two important steps. First we will convert all the given units, to international system. Later we will define the torque, which is given as the product between the radius of application of the force and the Force acting on the body. Mathematically the latter is,

$\tau = r F$

Here,

r = Radius

F = Force

Now the units,

$r = 1.95cm = 1.95*10^{-2}m$

Replacing,

$\tau = (1.95*10^{-2})(43.5N)$

$\tau = 0.8483N\cdot m$

Therefore the torque that the muscle produces on the wrist is $0.85N\cdot m$

3 0

3 years ago

A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t

harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

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

Where $\vec{r}$ is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), $\vec{F}$ is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that $r$ (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

$35cm * \dfrac{1m}{100cm} = 0.35m$

Now using the torque formula:

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

$Torque = rFsin(\theta)}$

Where $\theta$ is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector $\theta = 90°$ , thus :