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# M1 Dynamics Question watch

1. I know that this should be a simple question, but I really am quite stuck!

(For those who have the Heinemann Textbooks this is Ex 5F Q14)

Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a particle B of mass 0.5kg. Particle B is initially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero, find the speed with which A begins to move.

(The answer at the back of the book is 1.04m/s)

Thanks
2. v² = u² + 2as
vb² = 2x9.8x(0.7-0.2)
vb = 3.13

maua + mbub = mava + mbvb

so as the 2 particles will move at the same velocity after the string becomes tight .. then:

maua + mbub = mav + mbv

0 + 0.5x3.13 = 1xv + 0.5xv
1.57 = 1.5 v
v = 1.04 ms-1
3. I have one more question:

(M1 Ex4E Q6b)

A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a force Y acting at an angle of 45 degrees to the horizontal. The coefficient of friction between the body and the plane is 0.2. Find Y when the body is on the point of slipping down the plane.

I have tried the following method to solve this problem:

Resolving up the plane gives:
F - mgsin20 - Ycos65 = 0
F - 19.6sin20 - Ycos65 = 0

Resolving perpendicular to the plane gives:
R - mgcos20 + Ysin65 = 0
R - 19.6cos20 + Ysin65 = 0

As F=0.2R,

0.2(19.6cos20 - Ysin65) - 19.6sin20 - Ycos65 = 0
Y(cos65 + 0.2sin 65) = 3.92cos20 - 19.6sin 20

Therefore Y=-6.15N (3sf.)

However, the back of the book says that is 3.67N

Could anyone see where I went wrong, and what the correct method is?

Thanks again
4. you've used 65 degrees instead of the stated 45.
5. (Original post by Goldenratio)
you've used 65 degrees instead of the stated 45.
I did that purposely. Y = 20 + 45 = 65 degrees away from the perpendicular distance of R (I think)

If you use 45 degrees instead, the answer comes out to be -5.00N, which is still incorrect.

Perhaps a diagram would help
6. Anyone else?
7. (Original post by ebinsc)
I have one more question:

(M1 Ex4E Q6b)

A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a force Y acting at an angle of 45 degrees to the horizontal. The coefficient of friction between the body and the plane is 0.2. Find Y when the body is on the point of slipping down the plane.

I have tried the following method to solve this problem:

Resolving up the plane gives:
F - mgsin20 - Ycos65 = 0
F - 19.6sin20 - Ycos65 = 0

Resolving perpendicular to the plane gives:
R - mgcos20 + Ysin65 = 0
R - 19.6cos20 + Ysin65 = 0

As F=0.2R,

0.2(19.6cos20 - Ysin65) - 19.6sin20 - Ycos65 = 0
Y(cos65 + 0.2sin 65) = 3.92cos20 - 19.6sin 20

Therefore Y=-6.15N (3sf.)

However, the back of the book says that is 3.67N

Could anyone see where I went wrong, and what the correct method is?

Thanks again
Limiting equilibrium
=> Friction = coefficient of friction x normal reaction
= 0.2R

PERPENDICULAR TO PLANE: R = 2gcos20+Ycos65
PARALLEL TO PLANE: 0.2R = 2gsin20+Ycos25
0.2(2gcos20+Ycos65) = 2gsin20+Ycos25
Y(cos25-0.2cos65) = 2gsin20-0.4gcos20
Y = (2gsin20-0.4gcos20)/(cos25-0.2cos65)
Y = (19.6sin20-3.92cos20)/(cos25-0.2cos65)
Y = 3.67N
8. (Original post by Widowmaker)
Limiting equilibrium
=> Friction = coefficient of friction x normal reaction
= 0.2R

PERPENDICULAR TO PLANE: R = 2gcos20+Ycos65
PARALLEL TO PLANE: 0.2R = 2gsin20+Ycos25
0.2(2gcos20+Ycos65) = 2gsin20+Ycos25
Y(cos25-0.2cos65) = 2gsin20-0.4gcos20
Y = (2gsin20-0.4gcos20)/(cos25-0.2cos65)
Y = (19.6sin20-3.92cos20)/(cos25-0.2cos65)
Y = 3.67N
Thanks

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