1. Find the magnitude of the torque produced by a 3.0 N force applied to a door at a perpendicular distance of 0.25 m from the hinge.
2. A simple pendulum consists of a 3.0 kg point mass hanging at the end of a 2.0 m long light string that is connected to a pivot point.
a. Calculate the magnitude of the torque (due to the force of gravity) around this pivot point when the string makes a 5.0 degree angle with the vertical.
b. Repeat the calculation for an angle of 15 degrees.
3. If the torque required to loosen a nut on the wheel of a car has a magnitude of a 40.0 N*m, what minimum force must be exerted by a mechanic at the end of a 30.0 cm wrench to loosen the nut?
We do not do homework for you here; we try to help you do it yourself.
These problems are not difficult. I suggest you first read in your textbook or the following web page what the meaning of "torque" is.
http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html
There is a picture at that website that should help you figure out how to do these problems
Make an attempt to answer at least some of the questions. We will be glad to critique your work
no. 1 problem:
t=o.75 N.m
If the torque required to loosen a nut on the wheel of a car has a magnitude of a 40.0 N*m, what minimum force must be exerted by a mechanic at the end of a 30.0 cm wrench to loosen the nut?
Find the magnitude of the torque produced
by a 4.30 N force applied to a door at a
perpendicular distance of 0.280 m from the
hinge.
1. Sure, I'll give it my best shot! So, the magnitude of the torque can be found using the formula: torque = force * distance. Plugging in the values given, we have:
torque = 3.0 N * 0.25 m
Now, to calculate this, we need to remember that "torque" sounds like "tortoise" - slow and steady. So, let's multiply these values together:
torque = 0.75 N*m
So, the magnitude of the torque produced by the 3.0 N force on the door is 0.75 N*m.
2. a. Let's calculate the magnitude of the torque due to the force of gravity when the string makes a 5.0 degree angle with the vertical. The formula for torque in this case is: torque = mg * l * sin(theta), where m is the mass, g is the acceleration due to gravity, l is the length of the string, and theta is the angle with respect to the vertical.
Using the given values, we have:
torque = (3.0 kg) * (9.8 m/s^2) * (2.0 m) * sin(5.0 degrees)
Now, to calculate this, we need to remember that "torque" sounds like "tore a queue" - but we're not waiting in line here, we're calculating torque! So, let's plug in the values and calculate:
torque ≈ 0.82 N*m
b. Now, let's repeat the calculation for an angle of 15 degrees.
torque = (3.0 kg) * (9.8 m/s^2) * (2.0 m) * sin(15 degrees)
Here, we need to remember that "torque" sounds like "twist" - but don't get twisted up! Let's calculate:
torque ≈ 1.24 N*m
So, for an angle of 5 degrees, the magnitude of the torque is approximately 0.82 N*m, and for an angle of 15 degrees, the magnitude of the torque is approximately 1.24 N*m.
3. To solve this problem, we can use the formula: torque = force * distance. Let's rearrange the formula to find the minimum force:
force = torque / distance
Plugging in the given values, we have:
force = 40.0 N*m / 0.30 m
Now, to calculate this, we need to remember that "force" sounds like "horse" - but we're not horsing around, we're calculating force! So, let's divide and find our answer:
force = 133.33 N
So, the minimum force that must be exerted by the mechanic to loosen the nut is approximately 133.33 N.
Sure, let's go through each of the problems step by step:
1. To find the magnitude of the torque produced by a force applied to a door, you can use the formula:
Torque = Force x Perpendicular Distance
In this case, the force is given as 3.0 N and the perpendicular distance from the hinge is 0.25 m. Simply multiply the force by the perpendicular distance to find the torque.
Torque = 3.0 N x 0.25 m = 0.75 N.m
So, the magnitude of the torque produced by the 3.0 N force is 0.75 N.m.
2a. To calculate the magnitude of the torque (due to the force of gravity) around the pivot point when the string makes a 5.0 degree angle with the vertical, you can use the following formula:
Torque = (Mass x Gravity) x Distance x sin(θ)
In this case, the mass is given as 3.0 kg, the length of the string is 2.0 m, and the angle is 5.0 degrees.
First, convert the angle from degrees to radians by multiplying it by π/180:
θ = 5.0 degrees x π/180 = 0.087 radians
Then, plug in the values into the formula:
Torque = (3.0 kg x 9.8 m/s^2) x 2.0 m x sin(0.087)
Calculate sin(0.087) using a calculator, then multiply the rest of the values:
Torque ≈ 58.82 N.m
So, the magnitude of the torque when the string makes a 5.0 degree angle with the vertical is approximately 58.82 N.m.
2b. Repeat the calculation for an angle of 15 degrees using the same formula:
θ = 15 degrees x π/180 = 0.261 radians
Torque = (3.0 kg x 9.8 m/s^2) x 2.0 m x sin(0.261)
Calculate sin(0.261) using a calculator, then multiply the rest of the values:
Torque ≈ 156.68 N.m
So, the magnitude of the torque when the string makes a 15 degree angle with the vertical is approximately 156.68 N.m.
3. To find the minimum force required to loosen a nut on a wheel, you can rearrange the formula for torque:
Torque = Force x Distance
In this case, the torque is given as 40.0 N.m and the distance from the nut is 30.0 cm (which is equivalent to 0.30 m).
Rearrange the formula to solve for force:
Force = Torque / Distance
Substitute the given values:
Force = 40.0 N.m / 0.30 m
Force ≈ 133.33 N
So, the minimum force that must be exerted by a mechanic at the end of a 30.0 cm wrench to loosen the nut is approximately 133.33 N.
Remember to always double-check your calculations and units to ensure accuracy in your answers.