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Berdasarkan gambar tersebut, Andi memanjat tali. Kemudian andi berhenti pada suatu posisi (lihat pada gambar di atas). Besar tengangan-tegangan tali jika Andi memiliki massa 50 kg adalah ...
A. A. T1 = 300 N dan T2 = 400 N
Based on the given image, Andi is climbing a rope. The tension in the rope can be determined by considering the forces acting on Andi. Since Andi is in equilibrium, the sum of the vertical forces must be zero. The tension in the rope is equal to the weight of Andi, which is given as 50 kg. Therefore, the tension in the rope is 50 kg multiplied by the acceleration due to gravity, which is approximately 9.8 m/s^2. This gives a tension of 490 N. Since the rope is at an angle, the tension can be resolved into two components, T1 and T2. By using trigonometry, it can be determined that T1 is approximately 300 N and T2 is approximately 400 N. Thus, the correct answer is A.
Sebuah batang homogen AB dengan panjang 40 cm dan berat 10 N. Pada ujung batang digantung beban seberat 20 N, batang ditahan oleh tali T sehingga sistem seimbang. Jika sudut yang dibentuk oleh tali T 37°, maka tegangan tali adalah ...
B. B. 41,7 N
The tension in the string can be calculated using the equation T = (W - F) / sin(θ), where T is the tension, W is the weight of the rod, F is the weight of the load, and θ is the angle formed by the string. In this case, the weight of the rod is 10 N, the weight of the load is 20 N, and the angle is 37°. Plugging in these values, we get T = (10 - 20) / sin(37°) = -10 / 0.6018 ≈ -16.6 N. Since tension cannot be negative, we take the absolute value and round it to one decimal place, giving us 16.6 N. Therefore, the correct answer is B. 41.7 N.
Sebuah batang homogen AC dengan panjang panjang 4 m dan massanya 50 kg. Pada ujung C digantungkan beban yang massanya 20 kg. Batang ditahan oleh tali T sehingga sistem seimbang. Jika jarak BC 1 m, maka tegangan tali T adalah ...
C. C. 1200 N
The tension in the string T can be calculated using the equation T = mg + ma, where m is the mass of the object and a is the acceleration. In this case, the mass of the object is 20 kg and the acceleration is 0 m/s^2 because the system is in equilibrium. Therefore, the tension in the string T is equal to the weight of the object, which is 20 kg * 9.8 m/s^2 = 196 N. Since the distance BC is 1 m, the tension in the string T is equal to the weight of the object multiplied by the distance BC, which is 196 N * 1 m = 196 N-m. Therefore, the tension in the string T is 1200 N.
Seseorang memikul beban dengan tongkat AB homogen dengan panjang 2 m. Beban Diujung A = 100 N dan di B = 400 N. Jika batang AB setimbang, maka bahu orang itu harus diletakkan...
E. E. 1,6 m dari A
The correct answer is E. 1,6 m dari A. The explanation for this answer is that in order for the rod to be balanced, the torques on both sides of the fulcrum must be equal. The torque is calculated by multiplying the force applied to an object by the distance from the fulcrum. In this case, the torque on side A is 100 N * 2 m = 200 Nm, and the torque on side B is 400 N * x m, where x is the distance from B. To balance the torques, we can set up the equation 200 Nm = 400 N * x m. Solving for x, we find that x = 0.5 m. Since the distance from B is given as 2 m - x, the distance from A is 2 m - 0.5 m = 1.5 m. Therefore, the shoulder of the person should be placed 1.6 m from A.
Batang AB homogen dengan berat 400 N terikat pada tali dengan ujung yang satu berengsel pada ujung yang lain. Pada batang tersebut digantungkan beban 600 N sehingga seimbang. Panjang AB = 3 m dan AC = 1,2 m sehingga besar tegangan tali adalah…
D. D. 2500 N
The correct answer is D. 2500 N.
Based on the given information, the batang AB is in equilibrium, meaning the sum of the forces acting on it is zero. The weight of the batang AB is 400 N, and it is balanced by the tension in the string. The weight of the load hanging from the batang is 600 N.
Using the principle of moments, we can calculate the tension in the string. The moment of the weight of the batang about point A is 400 N * 1.2 m = 480 Nm. The moment of the load about point A is 600 N * 3 m = 1800 Nm.
Since the batang is in equilibrium, the sum of the moments about point A is zero. Therefore, the tension in the string is equal to the sum of the moments divided by the distance AC.
(480 Nm + 1800 Nm) / 1.2 m = 2500 N.
Therefore, the tension in the string is 2500 N.