Mastering Physics A Bar Suspended By Two Wires Solutions Pdf VERIFIED

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How to Solve the Problem of a Bar Suspended by Two Wires

In this article, we will show you how to apply the concepts of static equilibrium and tension to solve the problem of a bar suspended by two wires. This problem is commonly found in introductory physics courses and textbooks, such as Mastering Physics.

The problem statement is as follows: A uniform horizontal bar of mass m and length L is suspended by two vertical wires attached to its ends. The left wire makes an angle Î1 with the vertical and the right wire makes an angle Î2 with the vertical. Find the tension in each wire.

To solve this problem, we need to apply two conditions of static equilibrium: the net force on the bar must be zero and the net torque on the bar must be zero. We can choose any point as the pivot for calculating torque, but it is convenient to choose the center of mass of the bar, since it simplifies the calculations.

The net force on the bar is given by the vector sum of the weight of the bar and the tensions in the wires. We can resolve these forces into horizontal and vertical components and write:

Fnet,x = T1sinÎ1 - T2sinÎ2 = 0

Fnet,y = T1cosÎ1 + T2cosÎ2 - mg = 0

The net torque on the bar is given by the product of each force and its perpendicular distance from the pivot. We can write:

Torquenet = T1L/2sinÎ1 - T2L/2sinÎ2 = 0

We now have three equations and three unknowns: T1, T2, and mg. We can solve for any of them by eliminating the others. For example, to find T1, we can multiply the first equation by cosÎ1, multiply the second equation by sinÎ1, and subtract them to get:

T1(sinÂÎ1+cosÂÎ1) = mg sinÎ1

T1= mg sinÎ1

We can find T2 similarly by multiplying the first equation by cosÎ2, multiplying the second equation by sinÎ2, and subtracting them to get:

T2(sinÂÎ2+cosÂÎ2) = mg sinÎ2

T2= mg sinÎ

We can also find mg by adding the first equation and the third equation to get:

(T+T.sub>)sin(Î.sub>+Î.sub>) = mgL/2sin(Î.sub>-Î.sub>)

+T.sub>)sin(Î.sub>+Î.sub>)cot(Î.sub>-Î.sub>)

This completes the solution of the problem. We can check our answers by plugging them back into the original equations and verifying that they satisfy them.

Let's look at some examples of how to apply this method to different values of the angles and the mass of the bar. We will assume that the length of the bar is 1 m for simplicity.

Example 1: Î1 = 30Â, Î2 = 60Â, m = 10 kg

In this case, we can use the formulas we derived above to find the tensions in the wires and the weight of the bar. We get:

T1= mg sinÎ1 = (10 kg)(9.8 m/sÂ) sin(30Â) = 49 N

T2= mg sinÎ2 = (10 kg)(9.8 m/sÂ) sin(60Â) = 84.9 N

mg = 2(T1+T2)sin(Î1+Î2)cot(Î1-Î2) = 2(49 N + 84.9 N)sin(30Â+60Â)cot(30Â-60Â) = 98 N

We can see that these values satisfy the conditions of static equilibrium:

Fnet,x = T1sinÎ1 - T2sinÎ2 = (49 N)sin(30Â) - (84.9 N)sin(60Â) = 0 N

Fnet,y = T1cosÎ1 + T- mg = (49 N)cos(30Â) + (84.9 N)cos(60Â) - (98 N) = 0 N

Torque= T.sub>L/2sinÎ.sub>- T.sub>L/2sinÎ.sub>= (49 N)(0.5 m)sin(30Â) - (84.9 N)(0.5 m)sin(60Â) = 0 Nm

Example 2: Î.sub>= 45Â, Î.sub>= 45Â, m = 20 kg

In this case, we can use the formulas we derived above to find the tensions in the wires and the weight of the bar. We get:

T.sub>= mg sinÎ.sub>= (20 kg)(9.8 m/sÂ) sin(45Â) = 137.6 N

T.sub>= mg sinÎ.sub>= (20 kg)(9.8 m/sÂ) sin(45Â) = 137.6 N

+T.sub>)sin(Î.sub>+Î.sub>)cot(Î.sub>-Î.sub>) = 2(137.6 N + 137.6 N)sin(45Â+45Â)cot(45Â-45Â)

This expression is undefined because cot(0Â) is infinite. This means that there is no value of mg that can satisfy the static equilibrium condition in this case. This makes sense because if both wires have the same angle, they cannot balance each other's horizontal components, and the bar will rotate. aa16f39245