Respuesta :
Please refer to the figure showing the free body diagrams.
We apply the law of conservation of momentum wherein the summation of momentum equals zero. Momentum is force times distance. Force equals mass times gravity, if mass is known. g=9.8 m/s2
Note: I will not indicate the units. Since all units are uniform and in SI, the answer would also be in SI units.
1. What is the force the left support exerts on the beam?
Take the right support to be the fixed point (F3=F4=0). The equation would become
[tex]-44(9.8)(5/3 + 5/3) + F4(5/3)-117(9.8)( \frac{5}{3} / 2)=0 [/tex]
F2 = 945.7 N
2. What is the force the right support exerts on the beam?
Take the point where the gymnast is standing (F1=0) to be the fixed point. The equation would become
[tex](945.7)(5/3)-(117)(9.8)(( \frac{5}{3} /2)+5/3)+F4(5/3+5/3)=0[/tex]
F4 = 387.1 N
3. How much extra mass could the gymnast hold before the beam begins to tip? Now the gymnast (not holding any additional mass) walks directly above the right support.
Take the the left beam to be the fixed point (F2=0). The equation would become
[tex](-117)(9.8)( \frac{5}{3} /2)-44(9.8)(5/3)+F4(5/3)=0[/tex]
F4 = 1004.5 N, this should be equal to the force exerted by the man so the beam won't tip.
The force exerted by man is,
[tex]F1 = (44)(9.8) = 431.2 N[/tex]
The excess force the gymnast could carry would be 1004.5 N - 431.2 N = 573.3 N. Converting this to mass,
Extra mass = 573.3 N ÷ 9.8
Extra mass = 58.5 kg
We apply the law of conservation of momentum wherein the summation of momentum equals zero. Momentum is force times distance. Force equals mass times gravity, if mass is known. g=9.8 m/s2
Note: I will not indicate the units. Since all units are uniform and in SI, the answer would also be in SI units.
1. What is the force the left support exerts on the beam?
Take the right support to be the fixed point (F3=F4=0). The equation would become
[tex]-44(9.8)(5/3 + 5/3) + F4(5/3)-117(9.8)( \frac{5}{3} / 2)=0 [/tex]
F2 = 945.7 N
2. What is the force the right support exerts on the beam?
Take the point where the gymnast is standing (F1=0) to be the fixed point. The equation would become
[tex](945.7)(5/3)-(117)(9.8)(( \frac{5}{3} /2)+5/3)+F4(5/3+5/3)=0[/tex]
F4 = 387.1 N
3. How much extra mass could the gymnast hold before the beam begins to tip? Now the gymnast (not holding any additional mass) walks directly above the right support.
Take the the left beam to be the fixed point (F2=0). The equation would become
[tex](-117)(9.8)( \frac{5}{3} /2)-44(9.8)(5/3)+F4(5/3)=0[/tex]
F4 = 1004.5 N, this should be equal to the force exerted by the man so the beam won't tip.
The force exerted by man is,
[tex]F1 = (44)(9.8) = 431.2 N[/tex]
The excess force the gymnast could carry would be 1004.5 N - 431.2 N = 573.3 N. Converting this to mass,
Extra mass = 573.3 N ÷ 9.8
Extra mass = 58.5 kg
1) The forces the left support exerts on the beam will be 945.7 N.
2)The force the right support exerts on the beam will be 387.1 N.
What is force?
Force is explained as the multiplication of mass and acceleration. Its unit is Newton. It is the external agent which helps to change the shape and size of an object.
The given data in the question is;
m₁ is the mass of a gymnast = 44 kg
m₂ is the mass of beam= 117 kg
L is the length =5 m.
1) The forces the left support exerts on the beam will be 945.7 N.
Assume that the right support is the fixed point (F₃=F₄=0). The equation would change to
[tex]-44 \times 9.81 (\frac{5}{3+\frac{5}{3} } )+F_2(\frac{5}{3} )-117\times 9.81 (\frac{5}{6} )=0 \\\\ \rm F_2= 947.5 n[/tex]
Hence the forces the left support exerts on the beam will be 945.7 N.
2)The force the right support exerts on the beam will be 387.1 N.
Assume that the fixed point is where the gymnast is standing (F1=0). The equation would change to;
[tex](945.7) \times (\frac{5}{3} -(117) \times 9.81 \times \frac{5}6} +\frac{5}{3} +F_4(\frac{5}{3}+ \frac{5}{3} )=0 \\\\ F_4= 387.1 N[/tex]
3)The extra mass could the gymnast hold before the beam begins to tip will be 58.5 N
Assume the left beam is the fixed point (F₂=0). The equation would change to
[tex]\rm F_1 = 44 \times 9.81 \\\\ \rm F_1 =431.2 N[/tex]
The excess force by the gymnast is found as;
[tex]\rm F_E = 1004.5 N - 431.2 N \\\\\ \rm F_E = 573.3 N.[/tex]
The extra mass is found as;
[tex]\rm m_E= \frac{F_E}{g} \\\\\ \rm m_E= \frac{573.3}{9.81} \\\\ \rm m_E= 58.5 \ Kg[/tex]
Hence the extra mass could the gymnast hold before the beam begins to tip will be 58.5 N
4) The force the left support exerts on the beam 516.3 upwards.
The force on the left support is found as;
[tex]\rm F_L = F_2 - F_1 \\\\\ \rm F_L = = 947.5-431.2 \\\\ \rm F_L = 51.6.3[/tex]
Hence the force the left support exerts on the beam 516.3 upwards.
5) The forces the right support exerted on the beam will be 947.5N
[tex]\rm F_R = F_2 - F_3 \\\\\ \rm F_R = = 947.5-0 \\\\ \rm F_R = 947.5[/tex]
Hence the forces the right support exerted on the beam will be 947.5N
To learn more about the force refer to the link;
https://brainly.com/question/26115859
