shouldnt the T(2)sin 30 for the sum of Y around the pin be negative instead of postive ?
@jeangast77812 жыл бұрын
Very instructive work ! Thanks from France. Please go on !
@INTEGRALPHYSICS2 жыл бұрын
Thanks!
@gabifallous3175 Жыл бұрын
I’m not sure if this is a stupid question but how did you calculate the tension force? 3:12
@INTEGRALPHYSICS Жыл бұрын
In the numerator we have... 10+100 = 110 In the denominator 2sin(30) = 1 Put it all together and 110/100 = 1
@PatrickAndersson-pk5zd10 ай бұрын
What happens if the beam isn't in a horizontal position, but rather pointing 30 degrees upwards with the cable still holding it?
@INTEGRALPHYSICS10 ай бұрын
Fundamentally you find the solution following the same process; however you have to be careful with the angles surrounding the forces by gravity, they would no longer be 90 degrees.
@starvanza111 Жыл бұрын
how does it make sense that at that snapshot in time, when the beam is fully horizontal, there is a vertical normal force component exerted by the wall?
@INTEGRALPHYSICS Жыл бұрын
Something has to hold up the left side of the beam. Without that vertical component by the wall, the downward force by the weight of the beam would produce a net torque around the point where the cable attaches to the beam, allowing the beam to fall down.
@Rocker_Boy_86 Жыл бұрын
Can you please provide a diagram for the cable's fixing point on the wall? It's really confusing as there is the tension on both sides, the wall reaction etc..
@INTEGRALPHYSICS Жыл бұрын
A cable or string has equal forces on each end, however the forces are in opposite directions (both forces by the cable point inward toward the center of the cable). Meaning the force by the cable on the wall is down and to the right with a magnitude equal to the force by the cable on the beam.
@Rocker_Boy_86 Жыл бұрын
@@INTEGRALPHYSICS So there must be also a reaction force from the wall contradicting that end's tension, right?
@INTEGRALPHYSICS Жыл бұрын
Correct. The tension is down and to the right, the force by the wall is up and left.
@arcanestudio Жыл бұрын
By convention, shouldn't the first equation have (+) signs for counterclockwise torques and (-) signs for clockwise torques.? The equation that follows to solve for tension appears to be correct.
@INTEGRALPHYSICS Жыл бұрын
That is one conventions that irks me, not because it's wrong, but because it's completely arbitrary yet so many people teach it as an immutable law.
@arcanestudio Жыл бұрын
My comment was more along the lines of the absence of the (-) signs in the first equation that was presented rather than the sense of the torques. The sense really doesn't matter to me. That being said, both mechanical engineers and physicists seem to agree upon the same convention. With respect to the sense convention, it is similar to circular polarization: IEEE considers the sense of the departing polarization, while physicists consider the sense of the approaching polarization, the latter being contrary to the right-hand-rule.
@arcanestudio Жыл бұрын
There is mnemonic that is consistent with the cross product of two vectors that says that the vector points out of the page when curling the fingers counterclockwise with the right hand.
@INTEGRALPHYSICS Жыл бұрын
I do love the right hand rule.
@edilbertyu8657 Жыл бұрын
for the Sum of forces at Y is 50n Should be multiplied by the distance of 2m
@INTEGRALPHYSICS Жыл бұрын
No. That calculation is based on the sum of forces, on the system, not the sum of all torques(moments) about some point. Distance is not involved.
@EldarIusupzhanov Жыл бұрын
Why don't we calculate torque exerted by the beam using moment of inertia? I mean the torque of the beam = I*alpha = 1/3mr^2 * F/mr = 1/3*F*r ??
@INTEGRALPHYSICS Жыл бұрын
alpha (angular acceleration) is zero, the structure is not moving.