Explain why a negative sign is used before the 750(1.25)?

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Benjamin_harsh
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Homework Statement
Determine completely the resultant of the forces acting on the step pulley shown in this figure
Relevant Equations
##M_{axle} = 250 (1.25) + 1250(0.5) − 750(1.25)##
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##R_{X} = \sum F_{X}##

##R_{X} = 750.sin 60^{0} + 250##

##R_{X} = 899.52 N## to the right.

##R_{y} = \sum F_{y}##

##R_{y} = 750.cos 60^{0} - 1250##

##R_{y} = - 875 N##

##R_{y} = 875 N ## downward

##R = \sqrt {R_{x}^{2} + R_{y}^{2}}##

##R = 1254.89 N##

## tan θ_{X} = \large \frac {R_{y}}{R_{x}}##

## tan θ_{X} = \large \frac {875}{899.52}##

##θ_{x} = 44.21^{0}##

##M_{axle} = \sum M_{center}##

##M_{axle} = 250 (1.25) + 1250(0.5) − 750(1.25)##

##M_{axle} = 0##

Thus, ##R = 1254.89 N## downward to the right at ##θ_{x} = 44.21_{0}## and passes through the axle.
 

Discussion

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Is there a question in this ? Ah, in the thread title !

What made you attribute the minus sign ?

Are you aware of angle, angular speed, angular acceleration and angular moment of inertia ?
 
CWatters said:
Insert missing word...

Torque (or force) is a ______ quantity.
negative quantity or scalar quantity.
 
what does "passes through the axle" means?
 
jbriggs444 said:
Torque and force are scalar quantities?! Are you sure?

Sorry, both are vector quantities but how does this answer my question?
 
Benjamin_harsh said:
Sorry, both are vector quantities but how does this answer my question?
The point is that a torque has more than a magnitude. In two dimensions, it also has a sign -- clockwise or counter-clockwise. In three dimensions, it has a direction as well and can be treated as a vector.
 
In the diagram, Force 750 is acting in north east direction, It's X component will be +ve. But why negative sign is used?
 
BvU said:
Because it wants to make the wheel turn in a clockwise -- mathematically negative -- direction

Does this help ?

I thought clockwise means positive, so what makes clockwise direction is a negative thing?
 
Benjamin_harsh said:
I thought clockwise means positive, so what makes clockwise direction is a negative thing?
The other two torques want to make the wheel turn counter-clockwise.

Regardless of whether we label clockwise as positive or negative, one torque still acts opposite to the other two.
 
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Benjamin_harsh said:
Thus, ##R = 1254.89 N## downward to the right at ##θ_{x} = 44.21^{0}## and passes through the axle.

What is the meaning of "downward to the right "? Why not "upward to the right"? How to decide it?
 
Your words, not ours ... :smile:

I suppose you assume the axis of rotation is horizontal ...

'to he right' isn't so easily described :rolleyes: (but you took the 250 to be positive, so that's the +x direction)
 
BvU said:
'to he right' isn't so easily described :rolleyes: (but you took the 250 to be positive, so that's the +x direction
My final answer is right or wrong?
 
Your answer appears correct to me.

However you should really start your answer by declaring that you define upwards, right and anticlockwise as positive. Don't leave it to convention or let the examiner guess.

Your answer for the resultant force has components

RX = 899.52N
RY = −875N

That means the resultant is to the right and downwards _because_ you defined up and right as positive.

As an exercise you could define downwards and left as positive. Then repeat all the equations and your answers then would be:

RX = -899.52N
RY = 875N

That would _still_ mean the resultant is to the right and downwards _because_ this time you defined downwards and left as positive.