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This is from OCR Mechanics 1:

A block of wood of mass 4kg is released from rest on a plane inclined at 30 degrees to the horizontal. ASsuming that the surface can be modelled as smooth (no friction), calculate the acceleration of the block, and its speed after it has moved 3m.

I'm assuming I have to resolve it parallel to the ramp, but I don't know what to do! Can someone please help!!!!

A block of wood of mass 4kg is released from rest on a plane inclined at 30 degrees to the horizontal. ASsuming that the surface can be modelled as smooth (no friction), calculate the acceleration of the block, and its speed after it has moved 3m.

I'm assuming I have to resolve it parallel to the ramp, but I don't know what to do! Can someone please help!!!!

Scroll to see replies

A block of wood of mass 4kg is released from rest on a plane inclined at 30 degrees to the horizontal. ASsuming that the surface can be modelled as smooth (no friction), calculate the acceleration of the block, and its speed after it has moved 3m.

4gsin30 = 4a

a = g/2

S = 3 u = 0 v = ? a = g/2 t = ?

v^2 = u^2 + 2as

v^2 = 3g

v = 5.42 ms^-1

Havent done m1 in a while so might be wrong .

4gsin30 = 4a

a = g/2

S = 3 u = 0 v = ? a = g/2 t = ?

v^2 = u^2 + 2as

v^2 = 3g

v = 5.42 ms^-1

Havent done m1 in a while so might be wrong .

insparato

A block of wood of mass 4kg is released from rest on a plane inclined at 30 degrees to the horizontal. ASsuming that the surface can be modelled as smooth (no friction), calculate the acceleration of the block, and its speed after it has moved 3m.

4gsin30 = 4a

a = g/2

S = 3 u = 0 v = ? a = g/2 t = ?

v^2 = u^2 + 2as

v^2 = 3g

v = 5.42 ms^-1

Havent done m1 in a while so might be wrong .

4gsin30 = 4a

a = g/2

S = 3 u = 0 v = ? a = g/2 t = ?

v^2 = u^2 + 2as

v^2 = 3g

v = 5.42 ms^-1

Havent done m1 in a while so might be wrong .

Just checked the answers and you are spot on! Thank you

insparato

A block of wood of mass 4kg is released from rest on a plane inclined at 30 degrees to the horizontal. ASsuming that the surface can be modelled as smooth (no friction), calculate the acceleration of the block, and its speed after it has moved 3m.

4gsin30 = 4a

a = g/2

S = 3 u = 0 v = ? a = g/2 t = ?

v^2 = u^2 + 2as

v^2 = 3g

v = 5.42 ms^-1

Havent done m1 in a while so might be wrong .

4gsin30 = 4a

a = g/2

S = 3 u = 0 v = ? a = g/2 t = ?

v^2 = u^2 + 2as

v^2 = 3g

v = 5.42 ms^-1

Havent done m1 in a while so might be wrong .

sorry to be a pain, but how did u get that first equation?

Simba

Yup, it has to be since the block is moving parallel to the plane. If we used F = ma perpendicular to the plane we would get a = 0, since the block is in equilibrium when we consider motion perpendicular to the plane.

Ah, cheers, I will rep you tomorrow when my rep power comes back

I'm now trying the next question, and Ive still not grasped it.

A car of mass 850kg is travelling with acceleration 0.3ms^-2, up a straight road inclined at 12 degrees to the horizontal. There is a force resisting motion of 250N. Calculate the magnitude of the driving force.

I have tried but I keep getting the wrong answer.

A car of mass 850kg is travelling with acceleration 0.3ms^-2, up a straight road inclined at 12 degrees to the horizontal. There is a force resisting motion of 250N. Calculate the magnitude of the driving force.

I have tried but I keep getting the wrong answer.

You need to find the force gravity has on the car as it goes up the hill; this is just mgsin(12) by resolving forces.

So the driving force = resistance to motion + force gravity has on the car + force needed to accelerate by 0.3m/s^2

=> F = 250 + 850(9.8)sin12 + 850*0.3

=>F = 2237

So the driving force = resistance to motion + force gravity has on the car + force needed to accelerate by 0.3m/s^2

=> F = 250 + 850(9.8)sin12 + 850*0.3

=>F = 2237

El Matematico

You need to find the force gravity has on the car as it goes up the hill; this is just mgsin(12) by resolving forces.

So the driving force = resistance to motion + force gravity has on the car + force needed to accelerate by 0.3m/s^2

=> F = 250 + 850(9.8)sin12 + 850*0.3

=>F = 2237

So the driving force = resistance to motion + force gravity has on the car + force needed to accelerate by 0.3m/s^2

=> F = 250 + 850(9.8)sin12 + 850*0.3

=>F = 2237

Thanks, that makes sense!

No problem The main part of these questions is to figure out which forces are acting, which way all the forces are acting, then youre usually ok

I;ve got stuck again, I think I've grasped the basics but I can't apply it to harder questions.

A hanging basket of weight 50N is held in equilibrium by two light inextensible strings. One string is attached to a fixed point A and this string makes an angle of 60 degrees with the vertical; the other string is attached to a fixed point B and this string makes an angle of theta degrees with the vertical. Given that the tension in the string attached to A is T and the tension in the string attached to B in 2T, find the values of theta and T.

Blimey thats a long winded question. some help would be much apreciated, mechanics is becoming the bane of my life.

A hanging basket of weight 50N is held in equilibrium by two light inextensible strings. One string is attached to a fixed point A and this string makes an angle of 60 degrees with the vertical; the other string is attached to a fixed point B and this string makes an angle of theta degrees with the vertical. Given that the tension in the string attached to A is T and the tension in the string attached to B in 2T, find the values of theta and T.

Blimey thats a long winded question. some help would be much apreciated, mechanics is becoming the bane of my life.

Hi, you need to resolve the forces both horizontally and vertically. This will give us 2 equations in two unknowns which we can solve;

Horizontally, you must have Tcos30 = 2Tsin(theta)

Vertically, you must have 50 = Tcos(60) + 2Tcos(theta)

You should be able to solve these now to find theta and T

Horizontally, you must have Tcos30 = 2Tsin(theta)

Vertically, you must have 50 = Tcos(60) + 2Tcos(theta)

You should be able to solve these now to find theta and T

El Matematico

Hi, you need to resolve the forces both horizontally and vertically. This will give us 2 equations in two unknowns which we can solve;

Horizontally, you must have Tcos30 = 2Tsin(theta)

Vertically, you must have 50 = Tcos(60) + 2Tcos(theta)

You should be able to solve these now to find theta and T

Horizontally, you must have Tcos30 = 2Tsin(theta)

Vertically, you must have 50 = Tcos(60) + 2Tcos(theta)

You should be able to solve these now to find theta and T

Yep that makes sense, cheers. However, I am struggling to solve these equations! I assume you solve them simultaneously, but I can't do it I feel stupid.

They are a little trickier than I expected. I think the trick is to re-arrange the 1st equation to get 0=... so take the LHS over to the right. Then divide one equation by the other like this;

0/50 = (2Tsin(theta) - Tcos(30))/(Tcos(60) + 2Tcos(theta))

now we can take T out as a factor and cancel them. This leaves us with

0 = (2sin(theta) - cos(30))/(cos(60) + 2cos(theta))

=> 2sin(theta) = cos(30)

=>sin(theta) = sqrt(3)/4

=> theta = 25.659 degrees

=> T = 1

I think thats right. There should be an easier way though

0/50 = (2Tsin(theta) - Tcos(30))/(Tcos(60) + 2Tcos(theta))

now we can take T out as a factor and cancel them. This leaves us with

0 = (2sin(theta) - cos(30))/(cos(60) + 2cos(theta))

=> 2sin(theta) = cos(30)

=>sin(theta) = sqrt(3)/4

=> theta = 25.659 degrees

=> T = 1

I think thats right. There should be an easier way though

El Matematico

I think thats right. There should be an easier way though

(This isn't really any different from what you did, except you complicated matters a lot for no real reason - dividing by the 2nd equation doesn't actually change the bit you're trying to solve).

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