Why is the wedge force positive?

[PhysicsCanuck]

Homework Statement

(a) A man applies a force of F = 300 N at an angle of 60.0° to the door, 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.

(b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?

My question: Why is the force of the wedge not negative?

My reasoning: The torque is directed clockwise (and would be a negative value), and since r would be positive, then Fsin(theta) should be negative, rendering the torque negative.

Relevant Equations

t = rF
t = rF sin(theta)

As per (b) in the above image, or easily solved with t=rFsin(theta), the perpendicular force is 260N.
When inputting that value into the equation for torque, the value for torque is 520Nm, as per t=2.00*300*sin60.

Because the wedge prevents the door from moving, the torque on/at the doorknob (in the counter-clockwise direction) is equal and opposite the torque from/at the wedge, thus the wedge torque is -520Nm.

t_wedge = -520Nm = 1.50*F*sin(theta)
Therefore, F*sin(theta) = -347N
and F = -347N/sin(theta)

Since the provided answer is F = 347, then sin(theta) is equal to -1, and thus theta is equal to -90.

But in my estimation, the wedge force should be (negative) -260N since the opposing force (the perpendicular force pushing counter-clockwise on the door) is (positive) +260N.

Any new way of looking at this and reconciling these opposing forces would be appreciated.

Thank you very much

 

[haruspex]

Homework Statement: (a) A man applies a force of F = 300 N at an angle of 60.0° to the door, 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.

(b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?

My question: Why is the force of the wedge not negative?

My reasoning: The torque is directed clockwise (and would be a negative value), and since r would be positive, then Fsin(theta) should be negative, rendering the torque negative.
Homework Equations: t = rF
t = rF sin(theta)

View attachment 252587

As per (b) in the above image, or easily solved with t=rFsin(theta), the perpendicular force is 260N.
When inputting that value into the equation for torque, the value for torque is 520Nm, as per t=2.00*300*sin60.

Because the wedge prevents the door from moving, the torque on/at the doorknob (in the counter-clockwise direction) is equal and opposite the torque from/at the wedge, thus the wedge torque is -520Nm.

t_wedge = -520Nm = 1.50*F*sin(theta)
Therefore, F*sin(theta) = -347N
and F = -347N/sin(theta)

Since the provided answer is F = 347, then sin(theta) is equal to -1, and thus theta is equal to -90.

But in my estimation, the wedge force should be (negative) -260N since the opposing force (the perpendicular force pushing counter-clockwise on the door) is (positive) +260N.

Any new way of looking at this and reconciling these opposing forces would be appreciated.

Thank you very much

I suggest the question intends only asking for the magnitude.

 

[PhysicsCanuck]

I suggest the question intends only asking for the magnitude.

Hello and thank you,

Here is an explanation of why the sign matters, at the end of the solution.
I just don't understand why.

Notice that the angle from the position vector to the wedge force is -90°. That’s because, starting at the position vector, it’s necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle that way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition

 

[haruspex]

Hello and thank you,

Here is an explanation of why the sign matters, at the end of the solution.
I just don't understand why.

Notice that the angle from the position vector to the wedge force is 290°. That’s because, starting at the position vector, it’s necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle that way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition

Not sure what you are saying. Is this your reasoning or are you quoting from a provided solution?

 

[PhysicsCanuck]

Not sure what you are saying. Is this your reasoning or are you quoting from a provided solution?

Provided solution in the textbook. That's how they justify the force being positive while maintaining the torque as negative.

NB. I just realized that by cutting and pasting, the -90 showed up as 290 for some reason. I corrected it in my above comment, but it may not be showing in your response. Sorry about that

 

[haruspex]

Provided solution in the textbook.

They can't have it both ways. Both displacements are the same direction from the axis, so have the same sign. Both torques are calculated as $\vec r\times\vec F$. These must have opposite sign, so the forces must have opposite sign.

 

[PhysicsCanuck]

They can't have it both ways. Both displacements are the same direction from the axis, so have the same sign. Both torques are calculated as $\vec r\times\vec F$. These must have opposite sign, so the forces must have opposite sign.

Thank you

That is what I thought, but I wanted a professional opinion.

Have a great day