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PhyzicsOfHockey

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I don't understand these concepts very well. Could someone please take the time and help run through a few of these problems. I don't want the answers I want to know how to get them. Any help would be appreciated, thanks.

1. Two builders carry a sheet of drywall up a ramp.

Assume that Width = 2.10 m, Length = 3.10 m, Theta = 19.0°, and that the lead builder carries a (vertical) weight of 165.0 N (37.1 lb). What is the (vertical) weight carried by the builder at the rear?

I know the net Torq=0 since it is not rotating. I also know torque=r*F8sin theta. Other then that I don't know how to set up the problem. I don't know what I am suppose to set equal to each other.

2.A 33.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal.

If the angle between the beam and the cable is q = 62.0° what is the tension in the cable?

I took 1/2 the weight of the beam and divided it by the cos 62 and it gave me the right answer. I have no idea why this is. Could someone explain?

3.A 24.4 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 90o. If the beam is inclined at an angle of q = 22.2o with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)

I got Fh=1/2mgsin theta. Thats comes out to be 45.2 N and the answer is 41.87N. I think I am doing something wrong :-(

4.A window washer with a mass of 91.0 kg stands a distance, D = 0.600 m, from the left end of a plank of length, L= 2.30 m, with a mass of 15.0 kg. The plank is hung on two cables as shown. Find T2, the tension in the right cable.

For this one I did the Horizontal forces T1+T2=m1g+m2g

Also did torque1=T1(.6) and torq2=-T2(1.7) sooo T1(.6)=T2(1.7) T1=2.83T2

substituted 2.83T2+T2=m1g+m2g

3.83T2=m1g+m2g

T2=271.23N

But answer says its 301N, Haha yeah definitely not understaning something.

1. Two builders carry a sheet of drywall up a ramp.

Assume that Width = 2.10 m, Length = 3.10 m, Theta = 19.0°, and that the lead builder carries a (vertical) weight of 165.0 N (37.1 lb). What is the (vertical) weight carried by the builder at the rear?

I know the net Torq=0 since it is not rotating. I also know torque=r*F8sin theta. Other then that I don't know how to set up the problem. I don't know what I am suppose to set equal to each other.

2.A 33.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal.

If the angle between the beam and the cable is q = 62.0° what is the tension in the cable?

I took 1/2 the weight of the beam and divided it by the cos 62 and it gave me the right answer. I have no idea why this is. Could someone explain?

3.A 24.4 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 90o. If the beam is inclined at an angle of q = 22.2o with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)

I got Fh=1/2mgsin theta. Thats comes out to be 45.2 N and the answer is 41.87N. I think I am doing something wrong :-(

4.A window washer with a mass of 91.0 kg stands a distance, D = 0.600 m, from the left end of a plank of length, L= 2.30 m, with a mass of 15.0 kg. The plank is hung on two cables as shown. Find T2, the tension in the right cable.

For this one I did the Horizontal forces T1+T2=m1g+m2g

Also did torque1=T1(.6) and torq2=-T2(1.7) sooo T1(.6)=T2(1.7) T1=2.83T2

substituted 2.83T2+T2=m1g+m2g

3.83T2=m1g+m2g

T2=271.23N

But answer says its 301N, Haha yeah definitely not understaning something.

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