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Numerade Educator

# A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0$^\circ$ below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it ($\textbf{Fig. E11.20}$). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension $T$ in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

## $T = 7.64 \, \rm kN$, $\vec{P} = (3.2 \hat{i} - 0.5 \hat{j}) \, \rm kN$

#### Topics

Equilibrium and Elasticity

### Discussion

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LT

Ludvig T.

May 4, 2020

this is the worg question!

LT

Ludvig T.

May 4, 2020

this is the wrong question!

LT

Ludvig T.

May 4, 2020

this is the wrong question!

LT

Ludvig T.

May 4, 2020

LT

Ludvig T.

May 4, 2020

LT

Ludvig T.

May 4, 2020

##### Marshall S.

University of Washington

##### Farnaz M.

Simon Fraser University

##### Jared E.

University of Winnipeg

Lectures

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### Video Transcript

in this problem, we're gonna talk about static equilibrium. So what we need to remember is that the some of the forces acting on a body for it to be inside equilibrium must speak with 20 as well as the some off the torques on the body. So both must be equal 20 That's what we're going to apply. So what we have in our problem is a beam that is supporting a street land. And the beam itself is supported by cable represented here. The figure in red that makes a right angle with the beam and also a pivot, uh, representative here in black, okay, and are go is to find, given that the length off the beam is 4.5 m the distance between the pivot and the center of mass, the center of gravity. I'm sorry. Off the being is 2 m and the distance between the pivot and the place where the force the cable force is applied is 3 m. And also we have the information that attention on the cable is equal to five. I'm sorry. Not attention on the table. This is the force that the street land exerts on the beam downwards is equal to in magnitude. It's five killer Newton's. And I wrote I wrote it with a minus. Sign here because I'm setting up a coordinate system where the Y axis points upwards and the X axis to the right. Okay, so and the weight force is equal to mine is 1.4 kg Newtons the J direction. Okay. Ah, so now we can go on. And our goal is to find the magnitude of the tension and also the horizontal and vertical components of the force of the pivot exerts on the beach. I'm gonna call that force. Okay, so I'm going to start with the torque equation, so notice that I'm gonna draw a force diagram, provide a diagram on the beam. So we have first the gravitational force MGI acting at a distance of 2 m from the pivot. Then we have the attention force acting at a distance of 3 m from the bean. And we have the force f from the from the street lamp, but that is actually at a distance off 4.5 m from the pivot. Okay, um and also, uh, there is a force the pivot force That's pointing somewhere. I'm not sure where yet. We're gonna find that out. I'm just gonna, uh, right here somewhere. It's pointing somewhere basically and has magnitude P Ah. Now I can write the total talk So the total torque is equal to first noticed that the torque will always lie in the Z direction here in the Z axis by the right hand rule. Notice that this angle here, first of all, this single here is 25 degrees. This means that this angle here a 65 degrees. So is this angle here? So by the right hand rule, um, the torque caused by f and mgr counterclockwise. I'm gonna call it positive. So and the torque is equal to f. The magnitude of the torque is f times sign of data times the distance, So f l l is the total length on the beam. Time sign off. 65 pleasant g times D times the sign of 65 minus R T h h is the distance to the point where the tension is applied. I notice that here, uh, this, uh, the the angle is just 90. So the center 90 is one and I wrote it with a minus sign because by the right hand rule, it's pointing in the clockwise direction. And this must be put to zero. So T s f l sign 65 pleasant G de sign of 65 divided by H No. F is equal to 5 kg Newtons l is 4.5 m. Then that's multiplied by the sign of 65 plus M. G. That's the weight of 1.4 killing Newton's times D time sign of 65 divided by H and age is equal to 3 m. So T is equal to 7.64 Killer Newton's This is the magnitude of attention. Now notice that we have to find the force p to the pivot. Exerts, um So I'm gonna write t in the vector form so that we can find P notice that t is such that, uh, this angle here straight five. So this tangle here must also be 25 degrees so we can write t s minus T times sign of 85 I plus t times cold sign of 25 j eso using that T's 7.64 killer Newton's We obtain that tes minus 3.2 I plus 6.9 j Cuba Newton's This is the value of attention. T Now notice that the sum of all the forces that is the weight plus p plus t plus, um yes is equal to zero and we want to find P. So p is minus T plus f was W now T is minus 3.2. I was 6.9 j Yes, ISS minus five. Okay, I'm five j and ah W is minus 1.4 j. So this is equal to three points to I minus 0.5 j. So the horizontal component is 3.2 and I e forgot to right here. But this is in killing Newton's. The Realtor component is 3.2 and the vertical one is zero point. Mine is your point which concludes our exercise

Universidade de Sao Paulo

#### Topics

Equilibrium and Elasticity

##### Marshall S.

University of Washington

##### Farnaz M.

Simon Fraser University

##### Jared E.

University of Winnipeg

Lectures

Join Bootcamp