A roofer pulls a 25.0 kg bundle of shingles up the side of a roof that has a 29.0 degree pitch. She applies a force to the shingles, counteracting the 18.5 N of friction, to accelerate the shingles at a rate of 0.50 m/s2. Calculate the force she applied to the shingles.
4 years ago
Answered By Emily D
Free-body diagrams are your best friend! The words "COUNTERACTING the 18.5N of friction" tells us friction is in the opposite direction of our pulling force, P
If that diagram in the bottom left corner makes things more confusing, ignore it. It's just a trig explanation of why g// = g*sin(29)
Now that we have a FBD and we have the net force acting on the object, we can find our missing force, P
4 years ago
Answered By Emily D
Free-body diagrams are your best friend! The words "COUNTERACTING the 18.5N of friction" tells us friction is in the opposite direction of our pulling force, P
If that diagram in the bottom left corner makes things more confusing, ignore it. It's just a trig explanation of why g// = g*sin(29)
Now that we have a FBD and we have the net force acting on the object, we can find our missing force, P
$\Sigma F_x=18.5N+g\times\sin\left(29^{\circ}\right)-P$ΣFx=18.5N+g×sin(29?)−P
$\Sigma F_x=m\cdot a=25kg\cdot\frac{-0.5m}{s^2}=-12.5N$ΣFx=m·a=25kg·−0.5ms2 =−12.5N *** NEGATIVE because of our chosen coordinate system
$-12.5N=18.5N+\frac{9.81m}{s^2}\cdot25kg\cdot\sin\left(29^{\circ}\right)-P$−12.5N=18.5N+9.81ms2 ·25kg·sin(29?)−P
P = 149.9N [up the roof]
If this answer doesn't make sense, ask a follow-up question! I'll do my best to keep an eye on the help forum
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