mass of \( 1 \mathrm{Mg} / \mathrm{m}^{3} \). If \( \mu=0,1 \) and the maximum stress in the belt may not exceed 3 MPa , find the power that the belt can transmit at a belt speed of \( 23 \mathrm{~m} / \mathrm{s} \).

To find the power transmitted by the belt, we first need to calculate the tension in the belt using the formula for maximum stress, which is given by: \[ \sigma = \frac{T}{A} \] where \( \sigma \) is the stress (3 MPa), \( T \) is the tension, and \( A \) is the cross-sectional area. The area can be calculated from the mass density (\( 1 \, \text{Mg/m}^3 = 1000 \, \text{kg/m}^3\)) and the belt width and thickness (which are not provided, so we will denote them as \( b \) and \( h \)): \[ A = b \times h \] Given that the maximum stress is 3 MPa (which is \( 3 \times 10^6 \, \text{Pa} \)), we can express the tension as: \[ T = \sigma \times A = 3 \times 10^6 \, \text{Pa} \times (b \cdot h) \] Next, power transmitted by the belt can be calculated using: \[ P = T \times v \] where \( v = 23 \, \mathrm{m/s} \) is the belt speed. Substituting for \( T \): \[ P = (3 \times 10^6 \, \text{Pa} \cdot b \cdot h) \times 23 \, \mathrm{m/s} \] Thus: \[ P = 69 \times 10^6 \cdot b \cdot h \, \text{W} \] To find the actual power, we would need specific values for the width and thickness of the belt. However, with the parameters given, you can see how they influence the power transmission!