Tension in a string is which force

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Homework Statement If you have a string being pulled on each end by 60N, what is the total tension in the string? Homework Equations N/A[/B] The Attempt at a Solution The answer in my textbook says 60N, but it doesn't make a whole lot of sense to me (even though it's obviously right). If you're pulling both ends with 60 Newtons of force, why isn't it 120N of force, total, in the string? Why is it that the tension is only 60N? If one end were tied to a fixed object, and the string was pulled by 60N, then the total tension would also be 60, no? So then do you get the same result from pulling the other end rather than keeping it stationary? If one end were to pulled by 70N, and the other by 60N, would the total tension be 70? I find tension fairly confusing, and any help would very much appreciated. If you tied one end of the string to the wall, and pulled the other end with a force of 60 N, what is the tension in the string? The case where the string is being pulled with a force of 60 N from both ends does not result in a tension of 120 N, because forces have magnitude and direction. You can't simply add the two force magnitudes here and ignore that one force is acting in the opposite direction from the other. In order to determine tension in a string or a bar or whatever, it's best to draw a free body diagram before proceeding with the analysis. If one end were tied to a fixed object, and the string was pulled by 60N, then the total tension would also be 60 Suppose, you tie...

Homework Equations Accelerating force = F Resultant force = ma The Attempt at a Solution D is the correct option. Tension in string = T REsultant force = F - T = ma T = F - ma or if there is no friction, is the accelerating force the resultant itself? How do I proceed from here. Is the mass not constant? This is one of those questions where you have to know when to quit. It asks for the string segment with the greatest tension. Do you have any doubts regarding the segment you identified? actually, I don't know how to arrive at the answer. I know the equations I gave should apply here but how do I choose which one of the strings to choose? I can't give up. THere should be some principles which apply so that i can arrive at the answer. When I look at it, it seems that all the tension are the same. the same force causes them to move at the same speed, right? it seems that all the tension are the same. The rightmost block has a mass, does it not? It is being accelerated at a constant rate "a." The next block connected to it is also accelerating at the same rate. Label the blocks L to R as 1, 2, 3, 4, ... , N, however many there are. They have masses, "m 1, m 2, ... , m N. The "net" force on each block is then "am N. What you have to identify are where the net forces for each block are applied, and the string tensions necessary to apply them to achieve the acceleration. Give it a try. The rightmost block has a mass, does it not? It is being accelerated at a constant rate "a." T...

4:29 is only about the case where the yoyo is at the highest point where gravity is pointed downwards and centripetally, so we include it. When the yoyo is at its lowest point, gravitational force points downwards, which is opposite to the centripetal direction. So we have to subtract it from the tension to get the centripetal force. When the yoyo is at the leftmost or rightmost point, we don't include gravitational force because the gravitational force at that point is pointing tangentially to the circle, which is unrelated to the centripetal direction. 7:29) Since the force of gravity is always pointing downward (or pointing in vertical direction), it has no horizontal component if you're considering the points at 3 o'clock and 9 o'clock. Therefore, you don't have to include the force of gravity. If you're considering other points on the left or right, you need to include the horizontal component of the force of gravity. You only include the horizontal component of a force when solving these problems because only forces with some horizontal direction can affect the circular motion of the object. Hope that helps! David gave examples in vertical circles. There is also a video on horizontal circle. But what if the yo-yo is swinging at another angle to the horizontal? For example, if the yo-yo has a mas 5g, the string length is 0.5m and gravity is 9.8m/s^2, and it is swung at an angle of 36 degrees to the horizontal (as opposed to 90 degrees when the yo-yo is swinging vertic...

Normal Force Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure \(\PageIndex \) equal in magnitude and opposite in direction to the weight of the food \(w\). (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force \(N\) equal in magnitude and opposite in direction to the weight of the load. We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force and here is given the symbol \(N\). (This is not the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object’s weight if the object is on an incline, as you will see in the next example. COMMON MISCONCEPTION: NORMAL FORCE (N) VS. NEWTON (N) In this section we have introduced the quantity normal force, which is represented by the variable \(N\) This should not be confused with the symbol for the newton, which is also represented by the letter N. These symbols are p...

Two forces act on each ball hanging on the string: a force of gravity and tension of the string. The balls are also charged, so they repel one another with electric force. We determine its size using Coulomb's law. Both balls are at rest, so the net force must be zero. To satisfy this postulate, the vector sum of electric force and the force of gravity must be of the same size and opposite direction as the tension of the string. In the picture we can find two similar triangles ("green" and "purple"). From the similarities of the two triangles we can express the unknown size of electric force and thus the charge of the balls. Two forces act on each ball hanging on the string: the force of gravity \( \vec.\] \(m\,=\,0.5\,\mathrm\] • Electrostatics (33) • (L2) • (L3) • (L3) • (L2) • (L4) • (L4) • (L4) • (L4) • (L4) • (L3) • (L3) • (L3) • (L3) • (L3) • (L3) • (L3) • (L3) • (L4) • (L3) • (L4) • (L2) • (L3) • (L4) • (L4) • (L4) • (L2) • (L3) • (L4) • (L3) • (L4) • (L2) • (L4) • (L4) • Direct electric current and circuits (24) • (L3) • (L3) • (L3) • (L3) • (L2) • (L3) • (L2) • (L4) • (L4) • (L2) • (L3) • (L2) • (L3) • (L3) • (L3) • (L2) • (L2) • (L2) • (L2) • (L3) • (L2) • (L3) • (L3) • (L3) • Magnetic field (21) • (L2) • (L2) • (L3) • (L3) • (L3) • (L3) • (L4) • (L4) • (L4) • (L3) • (L3) • (L3) • (L4) • (L4) • (L2) • (L2) • (L3) • (L3) • (L2) • (L3) • (L3) • Alternating electric current and circuits (13) • (L2) • (L3) • (L3) • (L3) • (L3) • (L3) • (L3) • (L3) • (L3) • (L3) • (L4...

The force of tension is along the length of the rope, and it acts equally on objects at both ends – the tire and the branch. On the tire, the force of tension is directed upwards (because tension in the rope is holding the tire up) while on the branch, the force of tension is directed downwards (the tightened rope is pulling down on the branch). How to Find the Force of Tension

\( \newcommand\] Thus, we find that the tension in the string increases with the square of the speed, and decreases with the radius of the circle. Exercise \(\PageIndex\): Possible trajectories (in red) that the block will follow if the string breaks. An object is undergoing uniform circular motion in the horizontal plane, when the string connecting the object to the center of rotation suddenly breaks. What path will the block take after the string broke? • A • B • C • D Answer Example \(\PageIndex\) is speed. • The speed is larger if the radius of the curve is larger (one can go faster around a wider curve without skidding). • The speed is larger if the coefficient of friction is large (if the force of friction is larger, a larger radial acceleration can be sustained). Example \(\PageIndex\) has the dimension of speed. • The minimum velocity is larger if the circle has a larger radius (try this with a mass attached at the end of a string). • The minimum velocity is larger if the mass is bigger (again, try this at home!). Exercise \(\PageIndex\)). If you shortened the string, how would the minimum angular velocity (measured at the top of the trajectory) required for the ball to make it around the circle change? • It would decrease • It would stay the same • It would increase Answer Banked curves As we saw in Example 6.3.1, there is a maximum speed with which a car can go around a curve before it starts to skid. You may have noticed that roads, highways especially, are bank...

Imagine holding a something heavy (a physics textbook) in your hand with your arm straight down, by your side. Then raise your arm, still completely straight, up so it is at at an angle with you body. You will notice that it is much more difficult to hold the book in that position than if you let your arm hang straight down. It is the same principle. When we've attached the two new wires, does the original blue wire not provide any force to support the block anymore? I mean is it now just extending the downward force of the block to the two new wires and otherwise just hanging around? And if this is the case could you please explain why? I can set up and do a problem like this but I just don't understand why that original wire no longer supports the weight anymore in any way. Thanks. The original blue wire will still provide the same amount of force to the block, and it still has the same amount of tension on it, but is still transferring that force to the block. To put it another way, if you were to cut the blue wire, the block would fall because you removed the force that the blue wire supplies to the block. The fundamental reason behind this is that two mutually perpendicular vectors do not affect one another.Now, any vector can be resolved into two mutually perpendicular vectors, right? In this case T₁ᵪ and T₁ᵧ are those components.Now since T₁ᵪ is perpendicular to the 100N force acting downwards, it will not balance it, it will not add to it; it will not affect it in ...

See also: You are given a system that is at rest; you know the mass of the object, and the two angles of the strings. In this example problem, there are two strings, one with an angle of 25 degrees, and the other with an angle of 65 degrees, and a mass: 5 kilograms. Label the tension from the strings as T1 and T2, respectively. The first thing to notice is that because the system is at rest, the forces in the x and y directions balance each other out. We can now create two equations. We first create an equal for the horizontal forces. $$T_1\cos(25) = T_2\cos(65)$$ This is because the X-vector of each string equals Force $\cdot\cos(\theta)$. We now have to balance the vertical forces. $$T_1\sin(25) + T_2\sin(65) = F_g$$ $F_g$ is the only force we know. It equals mass $\cdot$ gravity, which is $5 \cdot 9.8 = 49$. So, we can make these equations: $$T_1 \sin(25) + T_2\sin(65) = 49$$ $$T_1 \cos(25) = T_2\cos(65)$$ So, by expressing $T_2$ in terms of $T_1$ and plugging it back into the first equation, we can find a value for $T_1$, then figure out $T_2$. This section is just solving the problem. We have no more physics left to do, just algebra. We start by isolating $T_2$ in the section equation. $$T_2 = T_1 \dfrac$$ Now, we have our final answer. The tension in string 1 is $20.708 N$, and the tension in string 2 is $44.419 N$.