How to calculate the balance of two forces
Publish: 2021-05-01 13:50:56
1. (1) The resistance (sliding friction, approximately equal to the maximum static friction) is 0.05 times of the vehicle weight, that is, Fmax = 0.05mg = 250N, which can not be pulled with 200N, so it is static friction. The two forces in the horizontal direction are balanced, that is, the actual static friction is f = f = 200N.
(2) in order to move in a straight line at a constant speed, the two forces of tension and sliding friction should be balanced, That is, the pull force should be 250N.
(3) when the horizontal pull force is 350n, the external force of the trolley = pull sliding friction = 100N, that is, the trolley will make uniform acceleration linear motion. (because the sliding friction f = u * n, in this problem, the vertical direction n = mg, so the sliding friction is unchanged.)
(2) in order to move in a straight line at a constant speed, the two forces of tension and sliding friction should be balanced, That is, the pull force should be 250N.
(3) when the horizontal pull force is 350n, the external force of the trolley = pull sliding friction = 100N, that is, the trolley will make uniform acceleration linear motion. (because the sliding friction f = u * n, in this problem, the vertical direction n = mg, so the sliding friction is unchanged.)
2. Collinear opposite direction
3. Easy to use methods are: force vector diagram method, draw the triangle of force, and use geometric relationship to solve. In addition, the orthogonal decomposition method, which is definitely taught by the teacher of the building owner, decomposes the force along two vertical directions, but for different problems, the decomposition may be different. For example, an object on an inclined plane is best decomposed along the direction of the inclined plane and the vertical inclined plane
it's helpful to remember the good comments. Please post again for new questions. No more answers here. Thank you
it's helpful to remember the good comments. Please post again for new questions. No more answers here. Thank you
4. When the balloon rises at a constant speed, it receives a balance force in the vertical direction, that is:
F = gtotal = g balloon + m person, g = 150 + 60 * 9.8 = 738n
F = gtotal = g balloon + m person, g = 150 + 60 * 9.8 = 738n
5. The original formula = 1 / [[3 [3 [3 [3 [3 + 3]] + 1 / [[5 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 + 3)] + 1 / [[5 [5 [5 [3 [3 [3 [3 [5 [5 [5 [5 [5 [5 [5 [5 [5 [5 [5 [3 [5 [3 [3 [3 [3 [3 [5 [3 [3 [3 [3 [3 [5 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3 [3])])])])])])]) 1 / (2 √ 3) - 1 / (2 √ 5) + 1 / (2 √ 5) - 1 / (2 √ 7) +... + 1 / (2 √ 47) - 1 / (2 √ 49)
= 1 / 2-1 / (2 √ 49) )
=1/2-1/14
=3/7
.
= 1 / 2-1 / (2 √ 49) )
=1/2-1/14
=3/7
.
6. 1. A vertical upward friction force equal to the gravity of an object when it is at rest or sliding at a constant speed on a vertical wall
2. When climbing up or down on a vertical pole or rope at a constant speed, or when it is still, it is subject to vertical upward friction equal to its own gravity
3. When an object is pushed horizontally on the horizontal ground, but not pushed, it is subject to the static friction force opposite to the thrust force, which is equal to the thrust force
when an object is subjected to a horizontal push (pull) force, the object moves at a constant speed, and the object is subjected to the opposite force, which is equal to the dynamic friction of the push (pull) force. If the push (pull) force changes again, the motion state of the object will change, but the dynamic friction force will not change.
2. When climbing up or down on a vertical pole or rope at a constant speed, or when it is still, it is subject to vertical upward friction equal to its own gravity
3. When an object is pushed horizontally on the horizontal ground, but not pushed, it is subject to the static friction force opposite to the thrust force, which is equal to the thrust force
when an object is subjected to a horizontal push (pull) force, the object moves at a constant speed, and the object is subjected to the opposite force, which is equal to the dynamic friction of the push (pull) force. If the push (pull) force changes again, the motion state of the object will change, but the dynamic friction force will not change.
7. (1) The resistance (sliding friction, approximately equal to the maximum static friction) is 0.05 times of the vehicle weight, i.e. Fmax = 0.05mg = 250N. It can not be pulled with 200N, so it is static friction. The two forces in the horizontal direction are balanced, i.e. the actual static friction is f = f = 200N
(2) in order to move in a straight line at a constant speed, the tension and sliding friction should be balanced, that is, the tension should be 250N
(3) when the horizontal pulling force is 350 n, the external force of the trolley = pulling force sliding friction = 100 N, that is, the trolley will make uniform acceleration linear motion Because, the sliding friction f = u * n, in this problem, the vertical direction n = mg, so the sliding friction remains unchanged.)
(2) in order to move in a straight line at a constant speed, the tension and sliding friction should be balanced, that is, the tension should be 250N
(3) when the horizontal pulling force is 350 n, the external force of the trolley = pulling force sliding friction = 100 N, that is, the trolley will make uniform acceleration linear motion Because, the sliding friction f = u * n, in this problem, the vertical direction n = mg, so the sliding friction remains unchanged.)
8. I found it. It's bullshit upstairs. Let me ask you a question. Where did you draw the friction when you talked about the schematic diagram of force? There's an experiment you can do. Put a box on the table with a rope on the upper edge of the left side and the lower edge of the right side, and then pull it with the same force. You can feel it. In theory, it will turn over, but the desktop has support. But sliding friction seems to be a little different, because you can't feel it's going to flip. Why? Its strength can be transmitted, you know? If the pulling force and the supporting surface are horizontal, you are not just applying the pulling force to the connecting point. The pulling force is transmitted to the whole object. If you translate the horizontal pulling force up, down, left and right, there is no difference in the effect of the force If you can't understand it, you can imagine that the tabletop is smooth without friction, and then the same horizontal force is applied at different heights, and the effect is the same)
(although that's true, it's still a little guilty, which seems to contradict the first experiment)
modification. I talked about this problem with my classmates a few days ago, and he and I said that it's a problem of torque, which they taught in the competition class, I didn't listen to it. I don't know much about it. PS: downstairs, it's said that it can be regarded as a particle. Not all cases can be regarded as the leader of a particle··
(although that's true, it's still a little guilty, which seems to contradict the first experiment)
modification. I talked about this problem with my classmates a few days ago, and he and I said that it's a problem of torque, which they taught in the competition class, I didn't listen to it. I don't know much about it. PS: downstairs, it's said that it can be regarded as a particle. Not all cases can be regarded as the leader of a particle··
9.
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