can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.

1 answer:

6 0

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

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Citrus2011 [14]

Algorithm of the Nios II assembly program.

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

<h3>The Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally, the program will be written using a cpulator simulator in order to attain best result.

We are to

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is