Question

remarks notice that the angle from the position vector to the wedge force is -90°. this is because, starting at the position vector, its necessary to go 90° clockwise (the negative angular direction) to get to the force vector. measuring the angle in this 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. figure (b) illustrates the fact that the component of the force perpendicular to the lever arm causes the torque. question to make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob? ○ closer to the hinge ○ closer to the doorknob practice it use the worked example above to help you solve this problem. (a) a man applies a force of ( f = 3.00\times10^{2} n ) at an angle of 60.0° to a door, ( x = 2.30 m ) from the hinges. find the torque on the door, choosing the position of the hinges as the axis of rotation. ( square ncdot m ) (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) wont open the door? ( square n ) exercise hints: getting started | im stuck! a man ties one end of a strong rope 7.51 m long to the bumper of his truck, 0.588 m from the ground, and the other end to a vertical tree trunk at a height of 3.05 m. he uses the truck to create a tension of ( 7.85\times10^{2} n ) in the rope. compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point. ( square ncdot m )

Explanation:

Step1: Calculate the torque for part (a)

The formula for torque is \(\tau = rF\sin\theta\).
Given \(F = 3.00\times10^{2}\space N\), \(r = 2.30\space m\), \(\theta=60.0^{\circ}\).
Substitute the values into the formula: \(\tau=(2.30)(3.00\times 10^{2})\sin60^{\circ}\).
Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866\), then \(\tau=(2.30)(3.00\times 10^{2})(0.866)\).
First, calculate \((2.30)(3.00\times 10^{2}) = 690\).
Then, \(\tau = 690\times0.866 = 597.54\space N\cdot m\).

Step2: Calculate the force for part (b)

For rotational equilibrium, \(\tau_{applied}=\tau_{wedge}\).
We know \(\tau_{applied}\) from part (a) (\(\tau = 597.54\space N\cdot m\)) and \(r_{wedge}=1.50\space m\).
Using the formula \(\tau = rF\) (since the angle between the force and the lever - arm for the wedge is \(90^{\circ}\), \(\sin\theta = 1\)), we can solve for \(F\).
\(F=\frac{\tau}{r}\), substituting \(\tau = 597.54\space N\cdot m\) and \(r = 1.50\space m\).
\(F=\frac{597.54}{1.50}=398.36\space N\).

Step3: Calculate the torque for the exercise

First, find the perpendicular distance \(d\) from the base of the tree to the line of action of the force.
The vertical distance between the two points of the rope is \(h=3.05 - 0.588=2.462\space m\), and the length of the rope \(L = 7.51\space m\).
Using similar - triangles or the concept of the lever - arm, the perpendicular distance \(d\) (lever - arm) from the base of the tree to the line of action of the tension force.
We can use the formula \(\tau = rF\sin\theta\). Another way is to find the perpendicular component of the position vector with respect to the force.
The perpendicular distance \(d\) (lever - arm) can be found using the ratio of similar triangles. The torque formula \(\tau = F\times d\).
The length of the rope \(L = 7.51\space m\), the vertical distance \(h = 2.462\space m\).
The torque \(\tau=(7.85\times 10^{2})\times\frac{2.462}{7.51}\times7.51\) (alternatively, using \(\tau = rF\sin\theta\), where \(r = 7.51\space m\) and \(\sin\theta=\frac{3.05 - 0.588}{7.51}\)).
\(\tau=(7.85\times 10^{2})(3.05 - 0.588)\)
\(\tau=(7.85\times 10^{2})(2.462)=1932.67\space N\cdot m\).

Answer:

(a) \(598\space N\cdot m\)
(b) \(398\space N\)
Exercise: \(1933\space N\cdot m\)