CE Board Exam Randomizer

Question 1

Belt Friction

To derive the belt friction equation, consider an infinitesimal element of the belt that subtends a small angle $ d\theta $ on the surface of a pulley. Tension forces on either side of this segment are:

$ T $: acting tangentially at one edge
$ T + dT $: acting tangentially at the other edge

These forces are separated by an angle $ d\theta $, so they each deviate from the central direction by $ \frac{d\theta}{2} $. A normal force $ dN $ acts radially inward, and the frictional force $ dF $ opposes the relative slipping motion.

In equilibrium (neglecting second-order small terms), summing forces in the tangential direction yields:

$$ dT = \mu \, dN $$

In the normal direction:

$$ dN = T \, d\theta $$

Substituting into the tangential equation:

$$ dT = \mu T \, d\theta $$

Separating variables:

$$ \frac{dT}{T} = \mu \, d\theta $$

Integrating from $ T_2 $ to $ T_1 $, and from $ 0 $ to $ \beta $:

$$ \int_{T_2}^{T_1} \frac{dT}{T} = \mu \int_0^\beta d\theta $$

$$ \ln\left( \frac{T_1}{T_2} \right) = \mu \beta $$

$${T_1 = T_2 e^{\mu \beta}} $$

This final equation relates the tensions on the tight and slack sides of a belt in contact with a surface over an angle $ \beta $ (in radians), where $ \mu $ is the coefficient of static friction.

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 1: – Diagram

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 2: – Diagram

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 3: – Diagram

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 4: – Diagram

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 5: – Diagram

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 6: – Diagram

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 7: – Diagram

Answer

See images:

Belt Friction | Statics of Rigid Bodies – Problem 8: – Diagram

Answer

Use the belt friction equation with the wrap angle in radians.

\[ \begin{aligned} \beta&=\pi\ \text{rad} \\ {T_2\over T_1}&=e^{\mu\beta}=e^{0.30\pi}=2.566 \\ T_2&=200(2.566)=513\ \text{N} \end{aligned} \] $\boxed{T_2=513\ \text{N}}$

Answer

Solve the belt friction equation for μ.

\[ \begin{aligned} \beta&=210^\circ\left({\pi\over180^\circ}\right)=3.665\ \text{rad} \\ {900\over300}&=e^{\mu(3.665)} \\ \mu&={\ln3\over3.665}=0.300 \end{aligned} \] $\boxed{\mu=0.300}$

Answer

The torque equals the tension difference times the drum radius.

PSAD - Statics / Belt Friction / Engr. Janclyde Espinosa (Clidez)

Refer to the assembly below. If the weight of the block is 500kg and the coefficient of friction between the rope and the drum is 0.3,

Which of the following most nearly gives the minimum force P (in Newtons) that must be applied to raise the block?

7860
7340
7420
7660

Which of the following most nearly gives the minimum force P (in Newtons) that must be applied to prevent the drum from falling?

3060
3480
2890
3220

Solution pending in psadquestions/q29.json.

Belt Friction

Problem:

Problem:

Problem:

Problem:

Problem:

Problem:

Problem:

Problem:

Problem (Basic belt tension ratio):

Problem (Coefficient from belt tensions):

Problem (Belt brake torque):

Exam Generator Problems

Question Bank: q29