# Calculating Tension + Length of String for Connected Particles

• MHB
• Shah 72
In summary, the block of mass 4kg is held on a rough slope that is inclined at 30 degree to the horizontal. The coefficient of friction between the slope and the block is 0.2. A light inextensible string is attached to the block and runs parallel to the slope to pass over a small smooth pulley fixed at the top of the slope. The other end of the string hangs vertically with a block of mass 1kg attached to the other end. The system is released from rest. The tension in the string is 10.6N. The block of mass 4kg reaches the bottom of the slope in 1.2 seconds and the other block has not yet reached the pulley. The lower bound for the length of the string
Shah 72
MHB
A block of mass 4kg is held on a rough slope that is inclined at 30 degree to the horizontal. The coefficient of friction between the slope and the block is 0.2. A light inextensible string is attached to the block and runs parallel to the slope to pass over a small smooth pulley fixed at the top of the slope. The other end of the string hangs vertically with a block of mass 1kg attached to the other end. The system is released from rest.
a) work out the tension in the string.
After 12s the block of mass 4kg reaches the bottom of the slope. The other block has not yet reached the pulley.
b) work out a lower bound for the length of the string giving your ans to 2 sf

Shah 72 said:
A block of mass 4kg is held on a rough slope that is inclined at 30 degree to the horizontal. The coefficient of friction between the slope and the block is 0.2. A light inextensible string is attached to the block and runs parallel to the slope to pass over a small smooth pulley fixed at the top of the slope. The other end of the string hangs vertically with a block of mass 1kg attached to the other end. The system is released from rest.
a) work out the tension in the string.
After 1.2s the block of mass 4kg reaches the bottom of the slope. The other block has not yet reached the pulley.
b) work out a lower bound for the length of the string giving your ans to 2 sf
I got the ans for (a) which is T= 10.6
I don't understand how to calculate (b)

Last edited:
$L > \dfrac{1}{2}at^2$

skeeter said:
$L > \dfrac{1}{2}at^2$
T= 10.6N, a=-0.6m/s^2, I get s=0.43

Shah 72 said:
T= 10.6N, a=-0.6m/s^2, I get s=0.43

s= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?

skeeter said:
s= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?
I calculated using t=1.2s

skeeter said:
s= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?
Iam sorry that was a typo error. t= 1.2 s not 12s

skeeter said:
s= 0.43 ?

you calculated that value using t = 12 seconds, or was that a typo in your original post?
Iam getting the ans of s=0.43 m but the ans in the textbook is 0.66m

Per my calculations, the 4kg mass moves a distance of 0.44 m.

If that is also the length of the incline, then the height of the incline is 0.22 m.

The only way I can explain the “text” answer is that the two were added, yielding 0.66 m, which doesn’t make much sense to me. That would essentially say the small mass is still at the bottom of the vertical edge of the incline when the large mass is at the bottom of the slanted part of the incline … go figure

skeeter said:
Per my calculations, the 4kg mass moves a distance of 0.44 m.

If that is also the length of the incline, then the height of the incline is 0.22 m.

The only way I can explain the “text” answer is that the two were added, yielding 0.66 m, which doesn’t make much sense to me. That would essentially say the small mass is still at the bottom of the vertical edge of the incline when the large mass is at the bottom of the slanted part of the incline … go figure
I agree with you. Thank you so much.

## 1. How do you calculate tension in a string for connected particles?

The tension in a string for connected particles can be calculated using the formula T = (m1 + m2)g, where T is the tension, m1 and m2 are the masses of the particles, and g is the acceleration due to gravity. This assumes that the string is massless and there is no friction.

## 2. What factors affect the tension in a string?

The tension in a string is affected by the masses of the connected particles, the acceleration due to gravity, and the length and stiffness of the string. Other factors such as air resistance and friction can also affect the tension.

## 3. How does the length of the string affect the tension?

The tension in a string is directly proportional to its length. This means that as the length of the string increases, the tension also increases. This is because a longer string has more surface area and therefore more weight, which contributes to the overall tension.

## 4. Can the tension in a string ever be negative?

No, the tension in a string cannot be negative. Tension is a force that acts in the opposite direction of the weight of the object. If the tension were negative, it would mean that the string is pulling the object towards it instead of supporting it.

## 5. What is the relationship between tension and the angle of the string?

The tension in a string is directly proportional to the sine of the angle between the string and the horizontal. This means that as the angle increases, the tension also increases. This relationship is described by the equation T = mg sinθ, where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the string.

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