Fr = 283N.[41o] + 283N.[0o]
X = 283*cos41 + 283 = 496.6 N.
Y = 283*sin41 = 185.7 N.
Fr^2 = X^2 + Y^2
Fr^2 = 496.6^2+185.7^2 = 281,096.05
Fr = 530.2 N.
Ff = u*mg = 0.283 * 264.6 = 74.9 N.
a = (Fr-Ff)/m=(530.2-74.9)/27=16.9 m/s^2
The magnitude of each force is 283 N, the
force on the right is applied at an angle 41�
and the mass of the block is 27 kg. The
coefficient of friction is 0.283.
The acceleration of gravity is 9.8 m/s2 .
27 kg
μ = 0.283
283 N
41� degrees
283 N
What is the magnitude of the resulting acceleration?
Answer in units of m/s2
1 answer