Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑ = mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Given Information:
Initial temperature of aluminum block = 26.5°C
Heat flux = 4000 w/m²
Time = 2112 seconds
Time = 30 minutes = 30*60 = 1800 seconds
Required Information:
Rise in surface temperature = ?
Answer:
Rise in surface temperature = 8.6 °C after 2112 seconds
Rise in surface temperature = 8 °C after 30 minutes
Explanation:
The surface temperature of the aluminum block is given by
Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.
After t = 2112 sec:
The rise in the surface temperature is
Rise = 35.1 - 26.5 = 8.6 °C
Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.
After t = 30 mins:
The rise in the surface temperature is
Rise = 34.5 - 26.5 = 8 °C
Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.