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I posted these earlier this morning along with 2 other problems. I've figured out the other two so I'm reposting these to clean things up a bit. If you can help with either problem please do, this assignment is due pretty soon.

1) A rod of length 68.0 cm and mass 2.00 kg is suspended by two strings which are 38.0 cm long, one at each end of the rod. The string on side B is cut. Find the magnitude of the initial acceleration of end B.

A diagram can be found here: http://capaserv.physics.mun.ca/msup...3a_1016full.gif

my workings:

net force = 0

A + B - mg = 0

and

net torque = 0

picking pivot to be at point A

LB - mgR = 0

so solving the torque equation to find

B = (9.8)(2)(0.34) / (0.68)

B = 9.8N

so A = 9.8N as well (solving the net force eqn)

I'm not even sure if I have that much done right, or even if i had to do that, I guess the length of the strings comes into play somewhere but I haven't used them yet, I need a lot of help with this one I think.

2) A 28.0 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of theta=13.1° with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam?

a graphic is here: http://capaserv.physics.mun.ca/msup..._beamhinge1.gif

so my workings

Ty + Fy - mgsin(90-13.1) = 0

Fx - Tx = 0

LTy - L/2 mgsin(90-13.1) = 0

Ty = (9.8)(28)(sin(76.9) / 2

Ty = 133.63 N

Tsin76.9 = 133.63N

T = 137.20

Tcos76.9 = Tx

Tx = 31.1 N

Fx - 31.1 N =0

Fx = 31.1 N

In (1), you have determined the forces (A and B) on the rod before the string is cut. Now when B is cut, the force at B becomes zero. At that instant, the force on A and the weight are the only forces acting on the rod.

From these unbalanced forces, you can find the acceleration of the CoM and from the unbalanced torque, you can find the angular acceleration about the CoM. From these 2 numbers you can get the initial accelration at the end B.

(2) : Instead of resolving forces horizontally and vertically, it's a lot easier if you resolve them parallel and perpendicular to the rod. This way you only have one force to resolve. Don't panic...that'll only make things worse.

but I don't think I really understand his advice, I've tried it for the first one and didn't get the correct answer and for the second one, I'm not sure how to do what he's saying. So please give me some advice about where to go with these.

1) A rod of length 68.0 cm and mass 2.00 kg is suspended by two strings which are 38.0 cm long, one at each end of the rod. The string on side B is cut. Find the magnitude of the initial acceleration of end B.

A diagram can be found here: http://capaserv.physics.mun.ca/msup...3a_1016full.gif

my workings:

net force = 0

A + B - mg = 0

and

net torque = 0

picking pivot to be at point A

LB - mgR = 0

so solving the torque equation to find

B = (9.8)(2)(0.34) / (0.68)

B = 9.8N

so A = 9.8N as well (solving the net force eqn)

I'm not even sure if I have that much done right, or even if i had to do that, I guess the length of the strings comes into play somewhere but I haven't used them yet, I need a lot of help with this one I think.

2) A 28.0 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of theta=13.1° with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam?

a graphic is here: http://capaserv.physics.mun.ca/msup..._beamhinge1.gif

so my workings

Ty + Fy - mgsin(90-13.1) = 0

Fx - Tx = 0

LTy - L/2 mgsin(90-13.1) = 0

Ty = (9.8)(28)(sin(76.9) / 2

Ty = 133.63 N

Tsin76.9 = 133.63N

T = 137.20

Tcos76.9 = Tx

Tx = 31.1 N

Fx - 31.1 N =0

Fx = 31.1 N

Goku replied with the following:Goku replied with the following:

In (1), you have determined the forces (A and B) on the rod before the string is cut. Now when B is cut, the force at B becomes zero. At that instant, the force on A and the weight are the only forces acting on the rod.

From these unbalanced forces, you can find the acceleration of the CoM and from the unbalanced torque, you can find the angular acceleration about the CoM. From these 2 numbers you can get the initial accelration at the end B.

(2) : Instead of resolving forces horizontally and vertically, it's a lot easier if you resolve them parallel and perpendicular to the rod. This way you only have one force to resolve. Don't panic...that'll only make things worse.

but I don't think I really understand his advice, I've tried it for the first one and didn't get the correct answer and for the second one, I'm not sure how to do what he's saying. So please give me some advice about where to go with these.

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