moments and forces
a uniform horizontal beam of weight 200N and length2.5m is freely hinged at one end to a wall. a rope is attached to the free end of the rope. the other end of the rope is attached to the wall at a point vertically above the beam so that the rope makes an angle of 30 degrees with the verticle. calculate the tension in the rope
Do you know where to start on the Q? Perhaps an idea on the drawing? Any equations that might be involved, perhaps what concepts might be used? ;-;
Hi
I did draw a diagram but I dont know how to use it to solve the question. I feel like you have to use moments equation: force * perpendicular distance to the pivot.
I did draw a diagram but I dont know how to use it to solve the question. I feel like you have to use moments equation: force * perpendicular distance to the pivot.
Original post by Callicious
Do you know where to start on the Q? Perhaps an idea on the drawing? Any equations that might be involved, perhaps what concepts might be used? ;-;
Original post by misslaiba
Hi
I did draw a diagram but I dont know how to use it to solve the question. I feel like you have to use moments equation: force * perpendicular distance to the pivot.
I did draw a diagram but I dont know how to use it to solve the question. I feel like you have to use moments equation: force * perpendicular distance to the pivot.
Upload the diagram you've drawn - I'd think about taking moments about the hinge as we don't want to find the forces there.
Because its a uniformly distributed load, Tributary length (how much of the beam is supported by the rope) would be half the beam.
Half the beam weighs 100N acting downwards in the vertical direction.
Use trig to find out the resultant diagonal force, you have been given one angle at the top (30*) the other is a right angle (90*)
Half the beam weighs 100N acting downwards in the vertical direction.
Use trig to find out the resultant diagonal force, you have been given one angle at the top (30*) the other is a right angle (90*)
hi
Original post by Muttley79
Upload the diagram you've drawn - I'd think about taking moments about the hinge as we don't want to find the forces there.
Original post by misslaiba
hi
OK great - now take moments -
clockwise moment, the weight =
anticlockwise moment, the Tension =
oh!
I thought that uniform means that 200N act downwards directly from the centre???
I thought that uniform means that 200N act downwards directly from the centre???
Original post by Moments
Because its a uniformly distributed load, Tributary length (how much of the beam is supported by the rope) would be half the beam.
Half the beam weighs 100N acting downwards in the vertical direction.
Use trig to find out the resultant diagonal force, you have been given one angle at the top (30*) the other is a right angle (90*)
Half the beam weighs 100N acting downwards in the vertical direction.
Use trig to find out the resultant diagonal force, you have been given one angle at the top (30*) the other is a right angle (90*)
Original post by misslaiba
oh!
I thought that uniform means that 200N act downwards directly from the centre???
I thought that uniform means that 200N act downwards directly from the centre???
You are right - ignore @Moments post. Your diagram is correct, although your 30 degree angle is rather large, but that doesn't matter.
Original post by Moments
Because its a uniformly distributed load, Tributary length (how much of the beam is supported by the rope) would be half the beam.
Half the beam weighs 100N acting downwards in the vertical direction.
Use trig to find out the resultant diagonal force, you have been given one angle at the top (30*) the other is a right angle (90*)
Half the beam weighs 100N acting downwards in the vertical direction.
Use trig to find out the resultant diagonal force, you have been given one angle at the top (30*) the other is a right angle (90*)
I (semi-)understand your logic, but it is not the approach that the OP will have been taught, nor generic. Let's stick to basic moment taking.
alright, I think I get it how to do it now.
thanks very much!
thanks very much!
Original post by Muttley79
OK great - now take moments -
clockwise moment, the weight =
anticlockwise moment, the Tension =
clockwise moment, the weight =
anticlockwise moment, the Tension =
wait can you explain.
Original post by misslaiba
alright, I think I get it how to do it now.
thanks very much!
thanks very much!
Original post by salimyasin10
wait can you explain.
The problem is solved by using the principle of moments and take the pivot where the hinge is positioned.
Sum of clockwise moments = Sum of anticlockwise moments
is the question an A-level question?
Original post by Eimmanuel
The problem is solved by using the principle of moments and take the pivot where the hinge is positioned.
Sum of clockwise moments = Sum of anticlockwise moments
Sum of clockwise moments = Sum of anticlockwise moments
Original post by salimyasin10
is the question an A-level question?
It could be.
k, all im worrying about is that it might be a gcse question.
Original post by Eimmanuel
It could be.
Original post by salimyasin10
k, all im worrying about is that it might be a gcse question.
Look at your specs to see if you are required to use any trigonometric ratio to solve the physics problems. If yes, then it can appear or else they can still give you such question but with other extra info.
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