a 10 kg weight is on an inclined plane. 10 ft long and 6 ft high. if there is no friction, what force parallel to the plane will prevent the body from slidin?
The gravity force component parallel to the plane is
M*g*sinA, where
sinA = 0.6.
A is the angle that the incline makes with the horizontal. It is 36.87 degrees.
M*g*sinA = 58.8 N
applied up and along the incline will prevent the body from sliding.
To find the force parallel to the inclined plane that will prevent the body from sliding, you need to calculate the gravitational force acting on the body along the inclined plane. Here's how you can do it step by step:
1. Determine the angle of the inclined plane. Given that the inclined plane is 6 ft high and 10 ft long, you can calculate the angle using the trigonometric function "sine." The sine of an angle is defined as the length of the opposite side divided by the length of the hypotenuse. In this case, the opposite side is 6 ft and the hypotenuse is 10 ft, so the sine of the angle (θ) is equal to 6/10 or 0.6. Take the inverse sine (sin^(-1)) to find the angle θ.
θ = sin^(-1)(6/10) ≈ 36.87 degrees
2. Calculate the gravitational force acting on the body along the inclined plane. The force due to gravity can be determined using the mass (m) of the body and the acceleration due to gravity (g). The weight (W) of an object is given by the formula W = m * g, where g is approximately 9.8 m/s^2. In this case, the mass of the body is 10 kg, so the weight will be:
W = m * g = 10 kg * 9.8 m/s^2 = 98 N
3. Resolve the weight into two components: one perpendicular to the inclined plane and the other parallel to the inclined plane. The component of weight parallel to the inclined plane is W_parallel = W * sin(θ). Using the value of θ calculated in step 1, we can find:
W_parallel = 98 N * sin(36.87 degrees) ≈ 58.92 N
Therefore, a force of approximately 58.92 N, acting parallel to the inclined plane, will prevent the 10 kg weight from sliding on the frictionless inclined plane.