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Need help with 2 problems

  • Thread starter Scirel
  • Start date
10
0
1.A 110 kg scaffold is 6.8 m long. It is hanging with two wires, one from each end. A 520 kg box sits 2 m from the left end. What is the tension in the left wire?(g = 9.8 m/s2)

For this one I tried using:

6.8 *t=460*9.8*4.8+110*9.8*6.8/2

and solving for t, but I for some reaosn get the wrong answer. What is wrong with my setup?


2.A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.48, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip?

For this one, I tried:

Nf*L/2*sin(90-theta)=Us*Nf*L/2*sin(theta)

and solving for theta, but this seems wrong as well. What is wrong with my setup?
 

Doc Al

Mentor
44,803
1,061
Scirel said:
1.A 110 kg scaffold is 6.8 m long. It is hanging with two wires, one from each end. A 520 kg box sits 2 m from the left end. What is the tension in the left wire?(g = 9.8 m/s2)

For this one I tried using:

6.8 *t=460*9.8*4.8+110*9.8*6.8/2

and solving for t, but I for some reaosn get the wrong answer. What is wrong with my setup?
Where did you get 460?


2.A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.48, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip?

For this one, I tried:

Nf*L/2*sin(90-theta)=Us*Nf*L/2*sin(theta)

and solving for theta, but this seems wrong as well. What is wrong with my setup?
Why is there an L/2 on both sides?
 
10
0
1.

OOPS! I`m such an idiot. That was supposed to be the mass. If 460 is repalced with 520, will the problem be correct?

2. I thought I needed to meansure the length from where mg affects the ladder, in the middle. Also I`m having trouble since the only value given is the coefficient of friction.
 

Doc Al

Mentor
44,803
1,061
Scirel said:
1. OOPS! I`m such an idiot. That was supposed to be the mass. If 460 is repalced with 520, will the problem be correct?
Yes.

2. I thought I needed to meansure the length from where mg affects the ladder, in the middle.
Correct. So the torque produced by the weight of the ladder will have an L/2. But what about the other torque?

Also I`m having trouble since the only value given is the coefficient of friction.
That's all you need. Everything else drops out.
 
10
0
Correct. So the torque produced by the weight of the ladder will have an L/2. But what about the other torque?

Wait.. is it just L?
 

Doc Al

Mentor
44,803
1,061
Scirel said:
Wait.. is it just L?
Yes. The other torque (produced by the normal force of the wall on the top of the ladder) will contain a factor of L, not L/2.
 
10
0
So, the equation:

Nf*L/2*sin(90-theta)=Us*Nf*L*sin(theta)


Will solve the problem?

Making this L/2*cos(theta)=Us*L*sin(theta)

then

1/2=Us*tan(theta)

then
1/(2*Us)=tan(theta) and finally:

arctan(1/(2*Us))=theta?



Also, just to be sure, is the answer to:

It takes a force of 90 N to compress the spring of a toy popgun 0.1 m to "load" a 0.14 kg ball. With what speed will the ball leave the gun?

this:

8.0178
 

Doc Al

Mentor
44,803
1,061
Scirel said:
So, the equation:

Nf*L/2*sin(90-theta)=Us*Nf*L*sin(theta)


Will solve the problem?

Making this L/2*cos(theta)=Us*L*sin(theta)

then

1/2=Us*tan(theta)

then
1/(2*Us)=tan(theta) and finally:

arctan(1/(2*Us))=theta?
Yes.



Also, just to be sure, is the answer to:

It takes a force of 90 N to compress the spring of a toy popgun 0.1 m to "load" a 0.14 kg ball. With what speed will the ball leave the gun?

this:

8.0178
Yes (in m/s). (Assuming it's fired horizontally.)
 

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