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Senior one physics problem, stress analysis The mass of the object is known to be m, and F and θ are known (1) If the object is uniform, find F (2) If other conditions remain the same, let's find f '
cosaF=m+sinaF*u
The length of the light rod is L = 0.4m, one end of which is connected with a small ball with mass of M = 0.5kg, which makes circular motion in the vertical plane. When the ball passes through the highest point at the speed of 0.4m/s and the lowest point at the rate of 0.6m/s, the force acting on the ball and its direction are calculated? What I want to ask is why the ball is attractive to the ball when it passes through the lowest point? Who can analyze it? And please solve the whole set of problems
When the ball is at its lowest point, because it is subject to centripetal force, you have to think, how does the centripetal force come from? The ball is under gravity at the lowest point, and then the direction of the centripetal force is toward the direction of the light thin rod. So the light rod must have the attractive force on the ball to keep the ball moving in a circle
An object with mass m is still on the horizontal plane, and the dynamic friction coefficient between the object and the horizontal ground is μ. Firstly, the object is pulled by a force F inclined to the right with an angle of θ with the horizontal direction. In order to make the object move uniformly along the horizontal ground, the range of force F is calculated. (given that the local acceleration of gravity is g, it can be considered that the maximum static friction force is equal to the sliding friction force)
Positive pressure n = mg fsin θ
Friction force F = μ n
Horizontal resultant force
F = fcos θ - (mg fsin θ) μ
Because there is acceleration, the resultant force is greater than 0
Fcosθ-(mg-Fsinθ)μ >0
F>mgμ/(cosθ+sinθμ)
Make the object on the horizontal ground, so the upward component of F cannot be greater than the gravity of the object, otherwise, the object will be lifted
Fsinθ
Physics_ The synthesis and decomposition of force and the equilibrium conditions A uniform metal ball, sandwiched between a smooth vertical plane and a smooth inclined plane with an inclination of 45 degrees, falls at a constant speed. The value of G in the inclined plane is g, and the dynamic friction coefficient between it and the horizontal ground is u 2. The weight of the object is 2kg. One end of the two light ropes AB and AC is connected to the vertical wall, and the other end is tied to the object. A pull force F with a direction of 60 degrees to the horizontal plane is applied to the object, and both ropes must be straightened. The range of the tension f is calculated
1. Isolate the ball first
When the ball is subjected to the wall horizontal force F, gravity g ', the slope to the ball elastic force N,
In the vertical balance direction:
N×cos45＝G'
For the inclined plane, it can be seen from the drawing that the ball speed / slope speed = Tg45
That is, the slope is also in balance
In the horizontal direction
N×cos45＝uN'
In the vertical direction, n '= g' + G
Simultaneous solution of equations
G '= UG / (1-u)
2. Make up a picture
Physical force analysis. Composition and decomposition of forces A uniform rope with mass of M is suspended at two points a and B on the horizontal ceiling, and the angle between the tangent line ad of the rope end and the horizontal ceiling is θ, and the tension at the lowest point C on the rope is calculated The process should be detailed thank you
Divide the rope into two and take half. The tension is the horizontal force. F '= fcos angle = mgcos angle / 2Sin angle = mgcot angle / 2, understand?
A body with a mass of 200kg is placed on an inclined plane with an inclination of 30 degrees. The component force of gravity along the inclined plane and perpendicular to the inclined plane is calculated. (g = 10N / kg) It can be seen from the geometric relationship that: G1=Gsin∝ G2=Gcos∝ By substituting the data, we can get the following results: G1 = GSIN ∝ = 200 x 10 x sin30 degrees N = 1000N G2 = GCOS ∝ = 200 x 10 x cos30 degree n = 1730n A:_______ . It can be seen from the geometric relationship that: G1=Gsin∝ G2=Gcos∝
Do you want to know why the action of force can be decomposed, that is, why the action of force satisfies the parallelogram rule? If so, you draw a parallelogram ABCD, assuming that the object is initially stationary at point a
When the component force F1 = F1 and the angle between the two components is 120 degrees, the resultant force is equal to the component force I just want to pull. What if the resultant force is greater than or less than the component force I'm a very stubborn person. I have to find out what I think. Although I know that the question I'm asking is really a piece of cake But just want to understand... Trouble you In fact, my problem is to say that a clothes hanging shunzi... In the case of continuous shunzi Is it only 120 degrees, more than 120 degrees, the CIS will break quickly I thought. It shouldn't break when it's over 120 degrees Isn't the resultant force greater than the difference between the two components and less than the sum of the two components
If the force is large, it will produce acceleration along the direction of the hand, and the object will move
If the force is small, it will produce acceleration in the backhand direction, and the object will move in the opposite direction
When the two forces form an angle of 120, the resultant force and the third force are in the same direction
There are two identical ropes. One is used to practice climbing rope on the cross bar, and the other is used to dry clothes. For two people of the same quality, a and B hold the vertical rope and hang in the air. B pulls down the middle point of the horizontal rope to break the rope, then () A. The pull of the rope to a is greater than the gravity of the nail B. The pulling force of rope to a is greater than that of rope to rope C. At the moment before B breaks the rope, the tension in the rope must be less than the gravity of B D. The angle between the two parts of the rope must be greater than 120 ° at the moment before breaking the rope
A. C. A is still in the air, and a is affected by the balance force. The tension of the rope and the gravity of a are a pair of balance forces. The tension of a by the rope is equal to the gravity of a; the mass of a and B is equal to that of a; a is suspended in the air, and the rope is not broken, and the tension of the rope is equal to the gravity of a; B pulls the middle point of the taut rope to break the rope, and the tension of the rope to B If it is greater than the pull force of rope to a, the gravity of a and B is equal, so the pull force of rope B must be greater than that of B at the moment before breaking the rope;
B. The pulling force of rope to a and that of rope to rope are a pair of mutual forces, equal, reverse and collinear, so C is wrong;
D. At the moment before breaking the rope, the tension is greater than the gravity, so the angle between the two parts must be greater than 120 ° so D is correct;
Therefore, D
What formula is used to directly find the formula of angle cosine theorem of triangle with any three sides Cosine theorem formula cosa = (b ^ 2 + C ^ 2-A ^ 2) / (2BC) Why is it not an angle
Cosine formula, you also have to have, in fact, is to formulate the relationship between the two sides of an arbitrary triangle and the number of the third sides corresponding to both sides. According to the theorem of congruent triangle, the cosine value of the angle of the corner side (given by the angle in the selected folder of the triangle's two sides) can determine a triangle. Therefore, it is a triangle with the third side, Thus, the corresponding relationship between the law and the third edge of the cosine of the edge angle can be determined
What are the formulas of cosine theorem and sine theorem of triangle
Cosine: square of B = square of a, square of C - 2accosb sine: A / Sina = B / SINB = C / sinc
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