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Backpack on a line

by itryphysics
Tags: backpack, line
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itryphysics
#1
Oct25-08, 07:45 PM
P: 114
1. The problem statement, all variables and given/known data

The two trees in the figure are 6.9m apart. A back-packer is trying to lift his pack out of the reach of bears. (a)Calculate the magnitude of the force F that he must exert downward to hold a 15 kg- backpack so that the rope sags at its midpoint by 1.5m . (b)Calculate the magnitude of the force that he must exert downward to hold a 15 kg- backpack so that the rope sags at its midpoint by 0.16m .




2. Relevant equations



3. The attempt at a solution
Distance between the trees =6.6 m

Half of distance between the trees = b =6.9/2= 3.45 m

sag at mid point = a = 1.5 m

Consider the vertical right anglled triangle with ' b ' as base, ' a ' as height and the half portion of rope as hypotenues ' h '

The hypotenues = h = sq rt [ b^2 + a^2 ] ,

h = sq rt [ 3.45^2 +1.5^2 ]

h =3.762 m

Now the hypotenues ( h = 3.762 m ) represents force Fand side 'a' represents weight 98 N

F/h =(15kg*9.8)/a

F /3.762 =147 /1.5 m

F =368.7 N

This is answer is not right. I dont know what im doing wrong :S Please guide me
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Mark44
#2
Oct25-08, 08:40 PM
Mentor
P: 21,258
Quote Quote by itryphysics View Post
1. The problem statement, all variables and given/known data

The two trees in the figure are 6.9m apart. A back-packer is trying to lift his pack out of the reach of bears. (a)Calculate the magnitude of the force F that he must exert downward to hold a 15 kg- backpack so that the rope sags at its midpoint by 1.5m . (b)Calculate the magnitude of the force that he must exert downward to hold a 15 kg- backpack so that the rope sags at its midpoint by 0.16m .




2. Relevant equations



3. The attempt at a solution
Distance between the trees =6.6 m

Half of distance between the trees = b =6.9/2= 3.45 m

sag at mid point = a = 1.5 m

Consider the vertical right anglled triangle with ' b ' as base, ' a ' as height and the half portion of rope as hypotenues ' h '

The hypotenues = h = sq rt [ b^2 + a^2 ] ,

h = sq rt [ 3.45^2 +1.5^2 ]

h =3.762 m
So far, so good.
[QUOTE=itryphysics;1929899]
Now the hypotenues ( h = 3.762 m ) represents force Fand side 'a' represents weight 98 N
[quote]
The weight is wrong. The mass of the pack is 15 kg, so its weight is 15 * 9.8 = 147 N. Your later work seems to use the right value.
Quote Quote by itryphysics View Post

F/h =(15kg*9.8)/a

F /3.762 =147 /1.5 m

F =368.7 N

This is answer is not right. I dont know what im doing wrong :S Please guide me
I get the same answer you did. My reasoning is that the right triangle has sides 1.5 m and 3.45 m. and hypotenuse ~3.76m. The length of the short leg of the triangle has to be in the same ratio to the downward force as the hypotenuse length is to the force along the rope. IOW 1.5/15 = 3.76/F, or F = 37.6 (in kg) (I'm not really dealing with forces just yet, just mass. Converting the number I got for F to a force, I multiplied by 9.8, and got 368.7 N (rounded).

You said you didn't think this was correct. What's the answer you believe is correct?


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