- #1
WillP
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ok, I have my assignment due by noon today and it is now 2:00am (I've been up most of tonight trying to get 4 questions done and they are all driving me insane) Of the 4, 3 are dealing with static Equilibrium So if you can Help with ANY of them I'll really appreciate it.
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/msuphysicslib/Graphics/Gtype20/prob13a_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/msuphysicslib/Graphics/Gtype24/prob19a_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
And this is wrong I know, I really don't know where I've gone wrong but I only have 3 more tries left on this question so I'm getting a little stressed out by it.
3) A sledgehammer with a mass of 2.40 kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is r_cm=0.570 m, and the moment of inertia about the center of mass is I_cm = 0.0360 kgm2. If the hammer is released from rest at an angle of theta=59.0 degrees such that H=0.489 m, what is the speed of the center of mass when it passes through horizontal?
A graphic: http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob34_sledgehammer.gif
I'm really not sure what to do with this one... I'm getting really tired and stressed and my concentration is slowly dying...
4) A solid door of mass 33.10 kg is 2.40 m high, 1.00 m wide, and 2.94 cm thick. If the edge of the door has a tangential speed of 45.7 cm/s, what is the rotational kinetic energy of the door?
I've already found the Moment of inertia for the door about an axis through it's hinges to be 11 kg*m^2 but now I'm not sure where to go with it?
I know I haven't explained my own workings very well, but like I said it's 2am here and I can't really concentrate very well. I really need help, I've been trying to figure out these last 4 questions for a couple days now, the rest of the assignment was easy compared to 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/msuphysicslib/Graphics/Gtype20/prob13a_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/msuphysicslib/Graphics/Gtype24/prob19a_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
And this is wrong I know, I really don't know where I've gone wrong but I only have 3 more tries left on this question so I'm getting a little stressed out by it.
3) A sledgehammer with a mass of 2.40 kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is r_cm=0.570 m, and the moment of inertia about the center of mass is I_cm = 0.0360 kgm2. If the hammer is released from rest at an angle of theta=59.0 degrees such that H=0.489 m, what is the speed of the center of mass when it passes through horizontal?
A graphic: http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob34_sledgehammer.gif
I'm really not sure what to do with this one... I'm getting really tired and stressed and my concentration is slowly dying...
4) A solid door of mass 33.10 kg is 2.40 m high, 1.00 m wide, and 2.94 cm thick. If the edge of the door has a tangential speed of 45.7 cm/s, what is the rotational kinetic energy of the door?
I've already found the Moment of inertia for the door about an axis through it's hinges to be 11 kg*m^2 but now I'm not sure where to go with it?
I know I haven't explained my own workings very well, but like I said it's 2am here and I can't really concentrate very well. I really need help, I've been trying to figure out these last 4 questions for a couple days now, the rest of the assignment was easy compared to these.
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