Physics lever questions for junior 3, physical levers for junior 3

The minimum force is the maximum force arm, and both F1 and F2 should be perpendicular to the force of OB (i.e., OB is the maximum force arm, F1 and F2 are in different directions), OA G object F1 OB F2 BO

f1=f2

B should be chosen, because when AO is in a horizontal position, the minimum force exerted at the B end should be perpendicular to the 0b direction, according to the lever principle F1*OB=G*OA, where G is the weight of the weight, because OB=OA, so F1=G.

When the OB end is horizontal, the lever is balanced, and the minimum force F2 exerted at the B end is also perpendicular to the OA direction, but the weight is always vertically upwards due to the force of gravity, that is, perpendicular to the OB direction, forming an angle with the OA direction (the angle is equal to the angle between the OB and the dotted line), then F2*OB=G*OA is obtained according to the principle of the lever'（oa‘=oa*cos

A should be chosen because when f1 is equal to g, the lever balances and bo is not horizontally parallel, and when f2 is greater than g, the lever moves slightly counterclockwise.

The two horizontal balances are not in the same position, one is that the AO side remains horizontal, while the other is the OB end remains horizontal, so there is no way to compare.

f1l1=f2l2

Since l1l2 is not known, it cannot be compared.

It only takes 50N to lift the A end, which means that the power arm is 4 times that of the resistance arm, the center of gravity is close to the fulcrum, and the B end is thicker.

If the iron rod is uniform, the center is in the geometric center of the iron rod, and the power arm is twice that of the resistance arm.

pwd=1234 Extraction code: 1234 Introduction: Junior high school physics high-quality materials**, suitable for teachers at all stages, daily tutoring of students, sprint for the high school entrance examination, and skill improvement learning.

According to the principle of parallel leverage, f1 l1 = f2 l 2

The tensile force of A to G can be found to be 10N

So the pressure is 50-10=40n

The force at point A fa oa = f ob

The force at point A fa=fob oa=30 5 15n=10nThe pressure of the weight on the horizontal ground n=the support force of the horizontal ground on the heavy object f1f1=n=g-fa=50n-10n=40nAnalysis: The lever theorem is used Power Power Arm = Resistance The resistance arm also has a two-force balance.

Hello, first of all, a lightweight wooden pole means that the weight of the lever is not considered, and the weight of the lever will be considered again in the third year of junior high school, and then the principle of lever parallelism f1 times l1 = f2 times l 2

In the figure, it is known that L1 is ob, L2 is OA, F1 is the find, F2 is 30N, and F1=10N is obtained, that is, the pulling force of A on the object.

So the pressure of the weight on the horizontal ground is 50N-10N=40N OAOA

According to F1L1=F2L2, the pull force of the lever on the weight = (5*30) 15 = 10, pressure = support force = g - pull force = 50-10 = 40