Question

a force of 1:50 newton is required to pull a cabinet across a floor at a constant velocity what will be the frictional force exerted on the cabinet​

Answer 1

Given:-

• Applied force = F = 1.5 N
• Velocity is constant

To find:-

• Frictional force = f = ?

Answer:-

According to the question, the cabinet is being moved at a constant velocity, which means acceleration = a = 0 m/s²

Refer to the attachment for FBD.

We will not consider the forces in vertical direction.

For horizontal direction,

from FBD, we can say that,

F - f = ma

Putting available values,

→ 1.5 - f = (m * 0)

→ 1.5 - f = 0

→ f = 1.5 N

Hence, the value of frictional force is 1.5 Newton Ans.

Extra knowledge:-

• Frictional force arises because of the interlocking of the irregularities present on the surfaces in contact.
• Friction always acts opposite to the direction of relative motion.
• Friction is of two types - Static friction and Dynamic / Kinetic friction. Dynamic friction includes sliding friction and rolling friction.
• Static friction > Sliding friction > Rolling friction
• The maximum value of static friction is called limiting friction.
• Static friction is a self adjusting force. If the applied force is less than or equal to limiting friction, then static friction = applied force and the body will be at rest.
• If applied force is more than limiting friction, body will be in motion, and kinetic friction will act in place of static friction.
• Maximum value of static friction / limiting friction = μₛN, where μₛ is coefficient of static friction, and N is normal reaction.
• Kinetic friction = μₖN where μₖ is the coefficient of kinetic friction and N is normal reaction.
• μₛ > μₖ.