the roof on a house rises 1 .00 m over a horizontal distance of 3.5 m. a 71 kg roofer stands on the roof. is the frictional force that keeps the roofer from slipping equal in magnitude to fparallel or fperpendicular? what is the magnitude of this force?
I'm pretty sure its fparallel, but how do you find fparallel when it gives you a distance? the formula i've been using is fparallel= mg sin(theta)
pinecone
To find the magnitude of the frictional force, you are correct that you need to calculate the frictional force parallel to the incline, denoted as "fparallel".
However, the formula you mentioned, fparallel = mg sin(theta), is used for finding the parallel force on an inclined plane. In this case, you have a roof, not an inclined plane. We need to use a different approach to find the magnitude of the frictional force.
Since the roof rises vertically, there is no slope, and the angle theta would be zero degrees. Therefore, sin(theta) would be zero, which means mg sin(theta) would also be zero. This suggests that the frictional force does not contribute to keeping the roofer from slipping.
Instead, the force that keeps the roofer from slipping is the perpendicular force, which is equal in magnitude to the normal force. The normal force is the force exerted by the roof on the roofer in a direction perpendicular to the roof's surface.
To find the magnitude of this force, we use the equation F = mg, where F represents the force exerted by the roof on the roofer. Therefore, the magnitude of the force is:
F = mg = (71 kg)(9.8 m/s^2) = 696.8 N
So, the magnitude of the force that keeps the roofer from slipping is 696.8 Newtons.
To find the frictional force acting on the roofer, you're correct that you need to determine the component of the force parallel to the roof's surface. In this case, the force parallel to the roof is actually the gravitational force acting on the roofer.
To calculate the magnitude of fparallel, you can use the formula: fparallel = mg sin(theta), where m is the mass of the roofer (71 kg), g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of inclination of the roof (which we can calculate using trigonometry).
In this question, it is given that the roof rises 1.00 m over a horizontal distance of 3.5 m. To find the angle theta, you can use the definition of the tangent function, where tan(theta) = opposite/adjacent. The opposite side in this case is the 1.00 m rise, and the adjacent side is the 3.5 m horizontal distance.
So, tan(theta) = 1.00 m / 3.5 m.
To find the magnitude of theta, you can take the inverse tangent (tan^-1) of both sides:
theta = tan^-1(1.00 m / 3.5 m).
Now that you have the value of theta, you can substitute it into the formula for fparallel:
fparallel = (71 kg) * (9.8 m/s^2) * sin(theta).
Once you calculate sin(theta) and substitute the values into the formula, you will get the magnitude of the frictional force (fparallel) that keeps the roofer from slipping.
friction has to be parallel to the roof.
Friction=mu*mg*cosTheta
where theta= arctan(1/3.5)
solve for theta, then cos theta. If you wish to avoid trig, if tantheta=1/3.5, then cosine=3.5/sqrt(1+3.5^2) check my triangle on that.
Dont forget about the portion of weight sliding down the plane... mgSinTheta