A 10kg trunk lies on a horizontal rough floor,the coefficient of friction between the trunk and the floor is 0.75. Calculate magnitude of force P which is necessary to pull the trunk horizontally if the P is applied horizontally and at 30 degrees above the horizontal.
upward force=Psin30
horizontal force=Pcos30
forcefriction=normalforce*mu
= mu(mg-Psin30)
horizontal force=friction solve for P
To calculate the magnitude of force P required to pull the trunk horizontally, we need to consider the frictional force acting on the trunk.
First, let's calculate the frictional force:
Frictional force (Ff) = coefficient of friction (μ) * normal force
The normal force (Fn) is equal to the weight of the trunk:
Fn = mass (m) * gravitational acceleration (g)
Fn = 10 kg * 9.8 m/s^2 = 98 N
Ff = 0.75 * 98 N = 73.5 N
Next, let's resolve force P into horizontal and vertical components:
Horizontal component of P (Ph) = P * cos(30°)
Vertical component of P (Pv) = P * sin(30°)
Since we want to pull the trunk horizontally, the vertical component of P should balance the vertical forces acting on the trunk. The only vertical force acting on the trunk is its weight.
Weight (W) = mass * gravitational acceleration
W = 10 kg * 9.8 m/s^2 = 98 N
Therefore, Pv = 98 N
To pull the trunk horizontally, the horizontal component of P should overcome the frictional force:
Ph = Ff
Now, let's substitute the values into the equation:
P * cos(30°) = 73.5 N
Divide both sides by cos(30°):
P = 73.5 N / cos(30°)
Using a calculator:
P ≈ 84.6 N
So, the magnitude of the force P necessary to pull the trunk horizontally is approximately 84.6 Newtons.
To calculate the magnitude of force P required to pull the trunk horizontally, we need to consider the forces acting on the trunk.
1. Weight (mg): The weight of the trunk is the force acting vertically downwards, given by the mass (m) of the trunk multiplied by the acceleration due to gravity (g). In this case, we have m = 10 kg and g = 9.8 m/s^2, so the weight is W = 10 kg * 9.8 m/s^2 = 98 N.
2. Normal force (N): The normal force is the force exerted by the floor on the trunk, perpendicular to the surface of the floor. In this case, since the trunk lies on a horizontal floor, the normal force is equal in magnitude and opposite in direction to the weight, so N = 98 N.
3. Frictional force (F_friction): The frictional force is the force that opposes the motion of the trunk and is proportional to the normal force. The coefficient of friction (μ) between the trunk and the floor is given as 0.75. The formula for calculating the frictional force is F_friction = μN. So, F_friction = 0.75 * 98 N = 73.5 N.
Now, let's analyze the horizontal forces acting on the trunk when the horizontal force P is applied at 30 degrees above the horizontal.
4. Horizontal force (F_horizontal): This is the component of force P that acts horizontally and in the direction of motion. To find this force, we can use trigonometry. The horizontal component of P is given by F_horizontal = P * cos(30°).
Since the trunk is not moving vertically, the vertical component of P is balanced by the normal force N. Therefore, we do not consider it in this calculation.
Considering the equilibrium of forces, the horizontal force P should counteract the frictional force F_friction. So we set F_horizontal = F_friction:
P * cos(30°) = 73.5 N
To solve for P, divide both sides of the equation by cos(30°):
P = 73.5 N / cos(30°)
Using a calculator, we can determine:
P ≈ 84.6 N
Therefore, the magnitude of force P necessary to pull the trunk horizontally, applied at 30 degrees above the horizontal, is approximately 84.6 N.