# Browsing Tag: Force

hi friends and saw problem on a daily Mertz principle see what is given in problem find P required to acetate the block shown in figure below at two point five meter per second square that means no acceleration of box or block is given which is two point five meter per second square for this acceleration what should be magnitude of force P first step will draw Fb D of this block it will draw Fb D of block always in this D alembert’s principle Freebody diagram is very very important once will represent Freebody diagram applying delivers principle in one step we get solution of problem now what is weight of this block or mass of the block the mass of this block is given here as what 40 kg is mass which is given already in problem that means weight of this box even represent which is 40 into acceleration due to gravity G that is what nine point eight one what will get your normal reaction in upward direction that force P what is acting here is inclined at an angle of 15 degree with horizontal this one is 15 degree due to component of force P block definitely accelerates in right to a direction that means this is a direction of acceleration a if acceleration is a right toward inertia of force leftward mass is given as a 40 kg therefore 40 a as body accelerated right toward frictional force acts in leftward direction where this frictional force if I represent like this μ is given as a point three and this becomes RN this one is complete free body diagram now first we’ll find out here Rn applying equation along Y axis but RN direct is not possible which is in unknown for speak in terms of falsely summation of fy equal to 0 assuming say upward force is positive R an upward next one what will get upward component top P angle with horizontal is given vertical component becomes sine of this angle 15 degree then minus 40 G means 9.81 equal to 0 therefore this RN equal to this 14 to 9.81 is a 390 2.4 – this piece sine of 15° that means this is value of RF say equation number 1 next we’ll apply equation along x-axis summation of fx equal to 0 assuming rightward force is positive now component of P what we’ll get here P cos of 15 degrees frictional force minus 0.3 into RF and inertia force is what 40 into a equal to 0 now a is given in problem that is 2 point 5 meter per second square will substitute in next P cos of 15 minus 0.3 Rn value I’ll substitute from Equation number one three ninety two point four minus P sine of 15 minus 40 into isolation is two point five which is equal to 0 that means equation is in terms of only one unknown that is P it will calculate using calculator p value comes out to be two hundred and eight point six three Newton that means a problem is very easy because a problem is a single body problem therefore simply it will apply D alembert’s principle easily solution is possible thank you

So far, we have been dealing with concentrated, or point forces. We depict a concentrated force using an arrow. This arrow represents the magnitude and the direction of the force, and we apply the force at a specific point. In real life, forces are not always concentrated and not applied at a point. Consider this bookcase. Here the books on the shelf create a distributed force along the shelf. In many such instances, loads are distributed along lines or over surfaces or over volumes. We call these forces as distributed forces. Look at this example. Sandbags are placed on the beam creating a distributed load over the beam’s entire length. The wind load on the highway signboard acts on the entire surface. Fluid pressure acts on the entire submerged area. Wind loads, water loads, people sitting in
a classroom and cars on a bridge are all examples of distributed loads. A distributed force is at any point is characterized by its intensity and the direction. So, there are two quantities, when we deal
with a distribution: a loading curve and the intensity of the force. The intensity of a force acting on a line
is defined by force per length unit. This is expressed in Newton per meter in SI the system, or pounds per foot in the US system. Now the question is how do we deal with distributed forces when solving problems. As you can imagine, it is convenient to replace the distributed load with an equivalent concentrated force because we know very well how to handle concentrated forces. Fortunately, this is relatively an easy two-step process. First, to convert the distributed load to
an equivalent concentrated load, we determine the total force. The total force is the area under the loading curve. It is easy to find the area when we deal with a uniform distribution like the one shown here. It is a rectangle. Finding the area of a rectangle is something we all know very well. Here the area is defined by multiplying the
length of the beam and the intensity of the force. Second, we need to figure out the location, where to apply this force. It turns out the force must pass through the centroid of the area under the curve. That’s it. Let us look at this example, where we have a uniformly distributed load on the 2-meter long beam. The intensity of the force is 200 Newtons
total force, and it is equal to 200 x 2=400 Newton. The centroid of this area is right in the
middle of the rectangle, so the force is applied at a distance of 1 m. Simple! Isn’t it? Let’s solve a few problems and get more
practice in the next video.

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a successful operation. We learn from our partners like the American
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the best of theirs and mix it up. The more you train, the better you become. We want to be the best always. We train for that. We love that.

Welcome to The Gaming Area Jump Force 1.18 Characters (DLC Included)

Today I will explain, Newton’s law of gravitation. Suppose we have 2 bodies (mass). Mass of this body is m1. Mass of this body is m2. As per Newton’s law of gravitation, If there are 2 bodies, Then both of them will have force of attraction. This body will have force in this direction. This body will have force in this direction. And then force F1 & F2, Suppose this is F1, this is F2, Then F1 & F2 will be equal, and will be equal to Where G is universal constant. And remains same any where in the universe. And “r” is the distance between these 2 bodies. Here, It is assumed that body mass, is concentrated here. And this body mass, is concentrated here. And distance between these 2 points is “r”. And both mass may be equal or different, But force will be equal. Now force is also equal to , mass into (X) acceleration. Suppose this is acceleration “a1”. So this F1 is equal to m1a1 also , and this also. This means this, and this are equal, then m1 will get cancel out. Only m2 will remain here. Similarly here force F2, will be equal to mass m2 into (X) acceleration a2. When this, and this are equal, Then m2 will cancel out. and we get a2 is equal to, So here there are 2 things. One is force, one is acceleration. Here also 2 things. Force F2 and acceleration a2. In both the cases force will be equal, whatever mass m1 & m2 are there. And force will be proportional to their masses, and inversely proportional to distance. Now forces are equal. But acceleration “a1” & “a2” are not equal. This is a1. This depends upon the value of m2. This means acceleration of this, is depending on mass of this. Similarly acceleration a2, is depending upon mass m1. This mass m1, decides the accleration of this body. Similarly this mass, decide the acceleration of this body. So if m1 is more, then acceleration of this body will be more, a2 will be more. Here I have shown 2 bodies again. They have force of attraction, which is equal. But I have shown these 2 bodies, looks like in horizontal position. Suppose, Instead of horizontal, they are like this. Stll , same force will come. This force will be, equal to this force. Suppose they are like this. Still force will be there. But now this force, will be in this direction. This force will be, in this direction. Both will be always toward other mass. Suppose this is like this. Then also force will be there. These 2 masses can be any where. Here, here, here, may be like this, Force will be there. And it will remain same. Here there was 2 masses. Here I have three. All 3 will attract to each other. Between this & this, Force will be like this. And these 2 forces will be equal. And will depend upon this mass and this mass. Similarly between these two, Force of attraction will be there, toward each other. And this force will depend upon mass of this, and mass of this m2. Similarly between these 2, force of attraction will be there. and will be function of m1 & m3. These 2 are equal. These 2 are equal. These 2 are equal. But this and this is not equal. Because this force depends upon m1 & m2. This force depends upon m1 & m3. So this force, will be different than this force, and will be different than this force. Now here suppose in between these two, I keep another mass. Then what will happen? Here there were mass m1 & m2. m1 & m2. This force will continue. It will not be affected by this mass. In addition to earlier force, ther will be 2 other forces. One between this mass & this mass. m1 & m3. And the 3rd one, between this m3, and m2. So one force m1 & m2, Another one between m1 & m3, 3rd one between m3 & m2. This will not affect earlier force. Now here I have shown 1 very big mass. and one small mass. This very big mass may be earth also. So this mass, and this big mass will have force of attraction. So both will have force, it will be equal. Both will have acceleration also. But acceleration of this item, depends upon mass of other. Because this mass is more, so acceleration of this m2 will be more. So this will start moving this side, Because this mass is more. Similarly, suppose this, and this acceleration will be very less, about zero. Because this mass is very less compare to this. Now suppose this item is here. Then again it will start moving towards centre because of acceleration. But both will have force like this. Those force will be equal. I have not shown those force. I have shown only acceleration here. Now suppose this is here. Then again it will move toward centre, Because of force of attraction. Because this mass is much more than this. Similarly, because this mass is more, so acceleration of this is more. So this will move. Similarly suppose this mass is here. So again it will move towards centre. So this mass can be any where. It will start moving towards centre of the earth, because of force of attraction, Because of gravitational force. Now here again I have shown 2 mass, m1 and m2. They will have force of attraction towards each other. Now suppose this big body consist of 3 parts. This part, this part and this part. Then what will happen, This particle, this mass will have 3 forces. one for this, one for this, one for this, like this. This one, and this one will have force of attraction. This (C1) is centre mass of this. This (C2) is mass centre of this. This (C3) is the mass centre of this. So this one and this one will have force of attraction. This one, and this one again will have force of attraction. This one, and this one again will have force of attraction. But these 3 force together, will be equal to this force. Summation of these will not be different from this. It will be same. Because total item is same (one). Now here, this mass is having 3 forces. Suppose I move this mass from this point to inside. Say, like this. So this mass, and this mass will have force of attraction. Here I have not shown, there will be force. Similarly this mass and this will have force of attraction. This is mass centre of this. Similarly, mass centre of this portion is this. So there will be force of attraction between this m2 and this. Here again force will be there. I have not shown. Now here, this particle or this mass will have 3 forces. one this side. one this side, one this side. Earlier this was having like this. This force, total of these force will be more than this total of these forces. Because all was in same direction. But here this & this will get cancelled. So when we will add all these three, this force will be much less than this. So when we move some body, inside the earth, force of attraction or gravitational force comes down. This will reduce. Today we will close here it self.