a box at rest is in a state of equilibrium half way up on a ramp. the ramp has an incline of 42°. what is…
a box at rest is in a state of equilibrium half way up on a ramp. the ramp has an incline of 42°. what is the force of static friction acting on the box if box has a gravitational force of 112.1 n?\n70 n\n80 n\n75 n\n85 n
Answer
Answer:
A. 75 N
Explanation:
Step1: Resolve gravitational force
The gravitational - force component along the ramp is $F_{g\parallel}=mg\sin\theta$, where $mg = 112.1$ N and $\theta = 42^{\circ}$. $F_{g\parallel}=112.1\times\sin(42^{\circ})$
Step2: Calculate the value
Using a calculator, $\sin(42^{\circ})\approx0.6691$. Then $F_{g\parallel}=112.1\times0.6691\approx75$ N. Since the box is in equilibrium, the force of static friction $F_f$ balances the gravitational - force component along the ramp. So $F_f = F_{g\parallel}\approx75$ N.