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###### Find magnitude of third force

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1 year ago

Can someone simplify this question for me please?

From what I can understand. If two forces are pulling that would suggest that it’s under tension. The arrows suggest ones pulling to the left and one to the right so, force would be 8-6.5=1.5N but if I were to imagine two pulling forces, I would have said 8+6.5 as it accelerates uniformly for the first four seconds.

At four second, a third force pulls parallel, so it would act with the same acceleration? When it moves 2m same direction it’s speed is 1.24… it’s a little meh in my head.

Edit: as the velocity reduces after the new pulling force is added. I can make the assumption the third force is acting in the opposite direction of the original two.

From what I can understand. If two forces are pulling that would suggest that it’s under tension. The arrows suggest ones pulling to the left and one to the right so, force would be 8-6.5=1.5N but if I were to imagine two pulling forces, I would have said 8+6.5 as it accelerates uniformly for the first four seconds.

At four second, a third force pulls parallel, so it would act with the same acceleration? When it moves 2m same direction it’s speed is 1.24… it’s a little meh in my head.

Edit: as the velocity reduces after the new pulling force is added. I can make the assumption the third force is acting in the opposite direction of the original two.

(edited 1 year ago)

Reply 1

1 year ago

Original post by KingRich

Can someone simplify this question for me please?

From what I can understand. If two forces are pulling that would suggest that it’s under tension. The arrows suggest ones pulling to the left and one to the right so, force would be 8-6.5=1.5N but if I were to imagine two pulling forces, I would have said 8+6.5 as it accelerates uniformly for the first four seconds.

At four second, a third force pulls parallel, so it would act with the same acceleration? When it moves 2m same direction it’s speed is 1.24… it’s a little meh in my head.

From what I can understand. If two forces are pulling that would suggest that it’s under tension. The arrows suggest ones pulling to the left and one to the right so, force would be 8-6.5=1.5N but if I were to imagine two pulling forces, I would have said 8+6.5 as it accelerates uniformly for the first four seconds.

At four second, a third force pulls parallel, so it would act with the same acceleration? When it moves 2m same direction it’s speed is 1.24… it’s a little meh in my head.

When you use F=ma ... you need the NET force acting on the object.

There are two parts to the object's motion here:

(I) In the first four seconds, the object has some net force, hence it has some acceleration (given from F=ma) based on that net force (and the mass, of course). What is the velocity of the object at the end of this first phase of motion?

(II) There is a third force now acting on the object. The initial velocity for this phase of motion is the same as the velocity you found in part (I). You are told that the object displaces 2m in the same direction as its motion before (so, 2m RIGHTWARD). You also know that the final speed at the end of this phase is 1.24 m/s but this implies that the final velocity is either +1.24 or -1.24 ... both possibilities with lead to two possible forces F. In either case, you should introduce F acting rightward and determine what it is ... if it turns out to be negative, it means that it acts to the left instead. In order to find F, you need to use F=ma again .... what is the acceleration of the object in phase (II) ? Use the displacement, initial/final velocities in SUVAT to work that out.

EDIT: Based on the question's wording, I think they only want you to use +1.24 for the final velocity for (II). There is ambiguity in their phrasing when they say "moved a further 2m in the same direction" ... does that mean total distance travelled is 2m rightward, or does it mean the displacement from the end of phase (I) is 2m rightward?

Anyway ... choose +1.24 for first attempt and see what force you get.

(edited 1 year ago)

Since the two forces act in opposite directions, the net force acting on the object initially is indeed $F = 8-6.5 = 1.5{\text N}$. Assuming no resistive forces, the acceleration can be found with Newton's second law for constant mass ($F = ma$).

The third force, let's call it $G$, acts on the object after 4s, meaning the net force after 4s can be taken as $F = 1.5 + G ({\text N})$.

Next you need an equation of motion, one of your $suvat$ equations. Remembering that the initial velocity ($u$) is 0, you can break the motion into two parts by time [edit: and displacement!] (0-4s, 2m), characterised by two different accelerations. Ultimately, you are trying to find the acceleration in the second phase.

Let me know if anything's unclear.

The third force, let's call it $G$, acts on the object after 4s, meaning the net force after 4s can be taken as $F = 1.5 + G ({\text N})$.

Next you need an equation of motion, one of your $suvat$ equations. Remembering that the initial velocity ($u$) is 0, you can break the motion into two parts by time [edit: and displacement!] (0-4s, 2m), characterised by two different accelerations. Ultimately, you are trying to find the acceleration in the second phase.

Let me know if anything's unclear.

(edited 1 year ago)

Reply 3

1 year ago

Original post by GgbroTG

...you can break the motion into two parts by time (0-4s, 4-6s), characterised by two different accelerations...

Let me know if anything's unclear.

Let me know if anything's unclear.

My reading of the question suggests that the second phase involves the object moving 2m, not moving for 2s...

Original post by GgbroTG

Since the two forces act in opposite directions, the net force acting on the object initially is indeed $F = 8-6.5 = 1.5{\text N}$. Assuming no resistive forces, the acceleration can be found with Newton's second law for constant mass ($F = ma$).

The third force, let's call it $G$, acts on the object after 4s, meaning the net force after 4s can be taken as $F = 1.5 + G$ [${\text N}$].

Next you need an equation of motion, one of your $suvat$ equations. Remembering that the initial velocity ($u$) is 0, you can break the motion into two parts by time (0-4s, 4-6s), characterised by two different accelerations.

Let me know if anything's unclear.

The third force, let's call it $G$, acts on the object after 4s, meaning the net force after 4s can be taken as $F = 1.5 + G$ [${\text N}$].

Next you need an equation of motion, one of your $suvat$ equations. Remembering that the initial velocity ($u$) is 0, you can break the motion into two parts by time (0-4s, 4-6s), characterised by two different accelerations.

Let me know if anything's unclear.

As I theorised on my graph. I may be somewhat close. I should assume by the graph then one force is forward and the other is against.

for the first four seconds I can find the final velocity by finding the acceleration first from f=ma so, acceleration is 0.3260 and applying v=u+at to find final velocity 1.304

for the second stage the initial velocity is 1.304 and the final velocity is 1.24 over s=2 that way I find acceleration to be negative. By applying v²=u²+2as….

Allowing me to calculate 1.5+p₃=(4.4)(-0.04070704)

hence third force lol -1.1679 therefore it’s acting against

Original post by KingRich

Allowing me to calculate 1.5+p₃=(4.4)(-0.04070704)

hence third force lol -1.1679 therefore it’s acting against

Allowing me to calculate 1.5+p₃=(4.4)(-0.04070704)

hence third force lol -1.1679 therefore it’s acting against

Other than the mass being 4.6kg, as opposed to 4.4kg, this all seems correct (your final answer is slightly off as a result, mind). Good job.

(edited 1 year ago)

Original post by GgbroTG

Other than the mass being 4.6kg, as opposed to 4.4kg, this all seems correct (your final answer is slightly off as a result, mind). Good job.

Ahah, I have no idea why I keep writing 4.4! I must unknowingly hate the number 4.6 or something lol

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