Question

На перекладине АВ длиной 2 м и массой 10 кг повис на двух руках ученик Егорка массой 60 кг. Расстояние между вертикально расположенными руками 40 см. Определите силы давления перекладины на каждую опору, если известно, что правая рука ученика находится на расстоянии 60 см от опоры.

Answer 1

Правило моментов относительно левой оси
-F1*0+1*300+1*100+1,4*300-F2*2=0
F2=410 Н
правило моментов относительно правой оси
F1*2-1*300-1*100-0,6*300-F2*0=0
F1=290 Н

Answer 2

Calculation of Pressure Forces on the Supports

To determine the pressure forces exerted by the bar on each support, we need to consider the weight of the bar and the weight of the student hanging from it.

Given: - Length of the bar (AB): 2 m - Mass of the bar: 10 kg - Mass of the student (Egor): 60 kg - Distance between the vertically aligned hands: 40 cm - Distance of the student's right hand from the support: 60 cm

To solve this problem, we can use the principle of moments. The sum of the moments about any point must be zero for the system to be in equilibrium.

Let's consider the moments about the left support (A). The clockwise moments will be considered positive, and the counterclockwise moments will be considered negative.

Calculation Steps:

1. Calculate the total weight acting on the bar: - Weight of the bar: mass of the bar * acceleration due to gravity - Weight of the student: mass of the student * acceleration due to gravity

2. Calculate the moments about the left support (A): - Moment due to the weight of the bar: weight of the bar * distance from the center of mass of the bar to the left support (A) - Moment due to the weight of the student: weight of the student * distance from the center of mass of the student to the left support (A)

3. Set up an equation using the principle of moments: - Sum of clockwise moments = Sum of counterclockwise moments

4. Solve the equation to find the pressure forces on each support.

Let's perform the calculations:

Step 1: Calculate the total weight acting on the bar

- Weight of the bar: 10 kg * 9.8 m/s^2 = 98 N- Weight of the student: 60 kg * 9.8 m/s^2 = 588 N Step 2: Calculate the moments about the left support (A)

- Moment due to the weight of the bar: 98 N * (2 m / 2) = 98 N * 1 m = 98 N.m - Moment due to the weight of the student: 588 N * (2 m + 0.4 m - 0.6 m) = 588 N * 1.4 m = 823.2 N.m

Step 3: Set up an equation using the principle of moments

- Sum of clockwise moments = Sum of counterclockwise moments - 98 N.m - 823.2 N.m = 0

Step 4: Solve the equation to find the pressure forces on each support

- Pressure force on the left support (A): 0 N (since the left support does not exert any pressure force) - Pressure force on the right support (B): 98 N + 588 N = 686 N

Therefore, the pressure force exerted by the bar on the left support (A) is 0 N, and the pressure force exerted by the bar on the right support (B) is 686 N.

Please note that the calculations assume that the bar and the student are in equilibrium and neglect any other external forces or factors that may affect the system.

I hope this helps! Let me know if you have any further questions.