HELP (statics problem A-Level) please
The mark scheme for this question states that the resultant force should extend through the "on-ladder" words in the diagram. But on the diagram, the reaction at the wall was a longer arrow than the weight. (order of 1cm difference).
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
Unless I have been misunderstanding ∑F(x) = ∑F(y) = 0 for 2 years I don't see how the given answer is correct.
https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74081-QP-JUN17.PDF
It is question 5 on this paper I can't work out how to attach an image.
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
Unless I have been misunderstanding ∑F(x) = ∑F(y) = 0 for 2 years I don't see how the given answer is correct.
https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74081-QP-JUN17.PDF
It is question 5 on this paper I can't work out how to attach an image.
Reply 1
21
Original post
by ITS_AVAThe mark scheme for this question states that the resultant force should extend through the "on-ladder" words in the diagram. But on the diagram, the reaction at the wall was a longer arrow than the weight. (order of 1cm difference).
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
Unless I have been misunderstanding ∑F(x) = ∑F(y) = 0 for 2 years I don't see how the given answer is correct.
https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74081-QP-JUN17.PDF
It is question 5 on this paper I can't work out how to attach an image.
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
Unless I have been misunderstanding ∑F(x) = ∑F(y) = 0 for 2 years I don't see how the given answer is correct.
https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74081-QP-JUN17.PDF
It is question 5 on this paper I can't work out how to attach an image.
Is it the second part? If so, the wall reaction will be ~1/4 of the weight (simple moment estimate) so the floor reaction will be as they say, so roughly through the "on ladder".
Usually the lengths dont indicate magnitude, though making it longer than the weight is a bit misleading. If the floor reaction was "below the ladder", so making a clockwise angle > 30 with the vertical, then the system would not be in equilibrium (moments about the wall).
(edited 1 year ago)
Reply 2
Original post
by ITS_AVAThe mark scheme for this question states that the resultant force should extend through the "on-ladder" words in the diagram. But on the diagram, the reaction at the wall was a longer arrow than the weight. (order of 1cm difference).
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
I can understand your frustration with the "inaccurate" drawing of forces (mainly the length). In most of the examination boards, when the question does not state that the diagram is drawn to scale, it is advisable not to consider the length of the force to be the magnitude.
The description given by the MS is illustrated by the diagram below that I "steal" from ...
When the lines of action of the forces are extended, there is a common point of intersection for the three forces provided the system is in equilibrium. This is the main point that the board wants the students to appreciate.
If the resultant force from the ground is closer to the horizontal, the three forces will not intersect at a common point and the system cannot be in equilibrium.
Original post
by ITS_AVAThe mark scheme for this question states that the resultant force should extend through the "on-ladder" words in the diagram. But on the diagram, the reaction at the wall was a longer arrow than the weight. (order of 1cm difference).
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
Unless I have been misunderstanding ∑F(x) = ∑F(y) = 0 for 2 years I don't see how the given answer is correct.
https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74081-QP-JUN17.PDF
It is question 5 on this paper I can't work out how to attach an image.
This clearly implies that the upwards reaction at the floor, which equals the weight, will be less than the force of friction (which equals the reaction at the wall). So the resultant vector will be closer to the horizontal than the vertical.
Unless I have been misunderstanding ∑F(x) = ∑F(y) = 0 for 2 years I don't see how the given answer is correct.
https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74081-QP-JUN17.PDF
It is question 5 on this paper I can't work out how to attach an image.
For new users, TSR would restrict them from posting image files.
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