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An object of mass m glides along a rough slope at a constant speed_____ In addition to gravity, the resultant force of other external forces is equal to_____ , the direction is_______ .
The supporting force of gravity friction slope is vertical upward in three mg directions
The two wooden blocks with mass of M1 and M2 are connected with light thin wires. The horizontal pulling force F is used to pull the wood blocks to the right (1) How much tension does the thin wire exert on M1? (2) When the quality of two pieces of wood meets what conditions can we approximately think that the tensile force in the thin wire is equal to the external force F? Figure: left M1 --- line ----- M2 --- right pulling force F Of course there is friction
(1) Let the friction coefficient be w, the acceleration of the two blocks is a, and the sum of the friction forces on the block f = (M1 + m2) GW, then a = (F - (M1 + m2) GW) / (M1 + m2) = f / (M1 + m2) - GW, a + GW = f / (M1 + m2)
The weight of the object is 45 n, and the force with the size of F = 40 N and an angle of 60 degrees with the vertical direction is used to push the object upward, so that the object slides down the vertical wall at an even speed, and the dynamic friction coefficient between the object and the wall is calculated
Let the friction coefficient be u
45-fCOS60=ufSIN60
The solution is u = 5 √ 3 / 12
The weight of the object is 150 n, and the sliding friction coefficient with the horizontal ground is 3 / 2 of the root sign. The force F is applied obliquely upward at an angle of 30 ° with the horizontal direction. In order to make the object move uniformly along the horizontal direction to the right, calculate the value of the tension F Trouble gives the process
This is a very simple and typical problem of force balance force analysis
The coordinate system is established, and F is decomposed into two directions, and the forces in both directions are balanced,
G = fsin30 + F support
UF support = fcos30
Balance calculation of force A uniform wooden pole weighs 10N per meter, and its fulcrum is located at 0.3m away from the left end of the wooden pole. Now an object with a weight of 11n is hung on the left end of the wooden pole, which is set at the right end of the wooden pole. A vertical upward force of 5N is applied to make the wood pole balanced. How many meters is the length of the wooden pole? Where did your 0.3 * 10 * 0.3/2 come from on the first floor? He said that every meter weighs 10N... You will regard him as one meter And the gravity of the pole is g * (1 / 2l-0.3), isn't it
According to the moment balance, there are 11 * 0.3 + 5 * (l-0.3) + 0.3 * 10 * 0.3 / 2 = 0.3 * 10 * (l-0.3) / 2
You can solve it
The calculation of physical object's force in senior one In the figure, if the gravity of the two objects is GA = 20n, GB = 40n, friction factor between a and B is 0.2, and friction factor between B and ground is 0.4, after exerting force on B, relative sliding occurs between a, B, and ground, so as to calculate the friction force between each contact surface Isn't the friction between ab zero?
Now that the relative friction between them must have happened
A body is in equilibrium under the action of four common point forces. If one of the forces is 5 n What is the resultant force of the other three forces when they are rotated 90 ° anticlockwise? 5 times root number 2n
Instead of turning the force, we can first add a force of 5N opposite to the force, and then apply a force of 5N in the direction of 90 ° anticlockwise rotation, and the result is the same as that of turning
At this time, the resultant force is the original 0 plus the resultant force of the later two forces: root number (5 ^ 2 + 5 ^ 2) = 5 times root sign 2n
When a body is subjected to the following three forces in common, it is () A. 6N 8N 10N B. 1N 3N 5N C. 1N 8N 6N D. 6N 6N 6N
A. The maximum value of 6N and 8N is 14N, and the minimum is 2n. When 10N is taken, the minimum value is 0n, so a is correct;
B. When the resultant force is 5N, it is combined with the third force, and the minimum resultant force is not 0n, so B is wrong;
C. The combination of 1n and 8N has the maximum 9N and the minimum 7n, which can not be 6N, so it is impossible to balance with the third force, so C is wrong;
D. When 6N is taken as 6N, it may be balanced with the third force, so D is correct;
Therefore, ad
In order to keep the body in equilibrium, must the force acting on the body satisfy the common point force?
No, in order to keep the object in balance, only the resultant force of the force acting on the ground object is zero, but not necessarily a common point force. For example, when a vertical upward force is applied at both ends of a long wooden rod, the stick is in equilibrium, and the three forces (gravity and two vertical upward forces) on the wood rod are not in common
If a body is in equilibrium under the action of three forces, then the three forces are not parallel and have a common point
A body is in equilibrium under the action of several points in common An object is in a static state under the action of a number of common point forces, one of which is F1=35N, the direction is right, the other is F2=5N, the direction is left, and these forces are all in the same plane. Now, rotate F1 clockwise by 90 degrees, and F2 anticlockwise by 90 degrees. The magnitude of the resultant force on the object is
Except for F1 and F2, the resultant force of other forces is 30n, and the direction is horizontal to the left, which is balanced with the resultant force of F1F2. Now, the resultant force of new F1F2 after rotating F1F2 is 40 n, and the direction is vertical to the resultant force of other forces. Then the resultant force is 50 N!
RELATED INFORMATIONS
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