Problems Involving ForcesProblems Involving Forces

Frictional ForceFrictional Force

A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 14.0 N to keep the 40.0 N block moving at a constant velocity.

a) What is the coefficient of sliding friction for the block and the table?

b) If a 20.0 N brick is placed on the block, what force will be required to keep the block and brick moving at a constant velocity?

Frictional ForceFrictional ForceFN

W

FA

Ff

REMEMBER:

If the object is moving at a constant velocity, then the force applied (FA) will equal the Frictional force (Ff)

Given: EQN:

FA= Ff = 14.0 N Ff = µFN

W = -40.0 N

FN = 40.0 N

µ = ?

Part a: Ff = µFN

So µ = Ff/FN

µ = 14.0 N/40.0 N

µ = .350

Frictional Force Frictional Force REMEMBER:If the object is moving at a constant velocity, then the

force applied (FA) will equal the Frictional force (Ff)

Part B:

New weight:

W = -40.0 N + (- 20.0 N) = -60.0 N

Therefore, the Normal Force:

FN = 60.0 N

Constant Velocity makes:

FA = Ff

Ff = µFN so FA = µFN

FA = (.350)(60.0 N)

FA = 21.0 N

Force and AccelerationForce and Acceleration

A box weighs 75 N.

a) What is the mass of the box?

b) What is the acceleration of the box if an upward force of 90 N is applied?

Force and AccelerationForce and Acceleration

a) What is the mass of the box?

Given:

W = -75 N

g = -9.8 m/s2

W = mg → m = W/g

m = -75N/-9.8 m/s2

m = 7.7 kg

Force and AccelerationForce and Accelerationb) What is the acceleration of the box if an

upward force of 90 N is applied?

Since the question is asking for acceleration, we need to use Newton’s 2nd Law, Fnet = ma

Given:

W = -75 N

Upward force = 90 N

m = 7.7 kg

Fnet = ma → a = Fnet/m

a = (-75 N + 90 N)/7.7 kg

a = 15 N/7.7 kg

a = 1.9 m/s2