# Why is the wedge force positive?

• PhysicsCanuck
In summary, the wedge prevents the door from moving and the torque on the doorknob is equal and opposite the torque from the wedge, so the wedge torque is -520 Nm.
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

PhysicsCanuck said:
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.

haruspex said:
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

Last edited:
PhysicsCanuck said:
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?

haruspex said:
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

PhysicsCanuck said:
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.

haruspex said:
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

## 1. Why is the wedge force positive?

The wedge force is positive because it is a force acting in the same direction as the displacement of an object. In other words, when an object is being pushed or pulled along a surface, the wedge force is acting in the same direction as the object's motion.

## 2. How does the wedge force affect an object's motion?

The wedge force helps to accelerate an object in the direction of its motion. It also helps to overcome any opposing forces, such as friction, that may be acting against the object's motion.

## 3. Can the wedge force ever be negative?

No, the wedge force is always positive. Even if an object is being slowed down or stopped, the wedge force is still acting in the same direction as the object's motion.

## 4. What factors determine the magnitude of the wedge force?

The magnitude of the wedge force is determined by the force applied to the wedge, the angle of the wedge, and the coefficient of friction between the object and the surface it is being pushed along.

## 5. How is the wedge force related to the normal force?

The wedge force and the normal force are directly related. The normal force is the force exerted by a surface on an object that is in contact with it, and the wedge force is a component of this normal force acting in the direction of the object's motion.

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