How does using cosine in the Force of friction equation even impact this in any way? that still doesn't make any sense because the cos tells you the horizontal force acting on the block, which as i said in my first question does not effect the normal force, which wouldn't affect the friction
I don't really follow what your saying on the second part. The general equation the solution that i found is using i s
F = 0.1 * {(m*g)-10cos(10)}
Which comes out to an answer of 48N
Homework Statement
1st problem: If and object has a weight of 100 N and a frictional force of 50 N exerted on it, what is the coefficient of friction?
2nd problem: A Block with a mass of 50 kg, is pulled with a force of 10 N at an angle of 10 degrees to the horizon, if the coefficient of...