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 appriciate it.(adsbygoogle = window.adsbygoogle || []).push({});

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 dunno 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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# Homework Help: Running out of Time, Please help me !

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