Anonymous

# Help with physics question!!?

A 820 N crate is being pulled across a level floor by a force F of 400 N at an angle of 33° above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate.

Relevance

First find Fn

Sum Forces = Fg - Fn - Fay = 0

Fay = 217.86 N

Fn = 602.14 N

Then Find Ffr

Ffr = mu * Fn = .25 * 602.14 = 150.54

Then find a

Sum Fx = Fax - Ffr = ma

335.47 N - 150.54 N = 83.67 kg * a

a = 2.21 m/s^2

• Figure out the net force acting on the crate. Then use the famous equation "Fnet=ma" to solve for the acceleration.

In this problem, it's easiest to look at the vertical forces and horizontal forces separately.

Vertical forces:

a. Weight = mg = 820N (down)

b. Normal force from floor = F_n (up) [note we don't know how much that is yet.]

c. Upward component of the force due to the rope (I guess it's a rope). This is (400N)(sin33)

Since there is no acceleration in the vertical direction, this must mean the vertical forces are balanced (upward forces = downward forces). That is:

(1): F_n + (400N)(sin33) = 820N

Horizontal forces (assume crate is being pulled toward the right):

a. Horizontal component due to rope (pulling right). This is (400N)(cos33)

b. Friction (pulling left). this is (F_n)(μ)

The net force is the difference between the right-pulling force and the left-pulling force:

(2): Fnet = (400N)(cos33) − (F_n)(μ)

Now combine (1) with (2) to eliminate the variable F_n:

(3): Fnet = (400N)(cos33) − (820N−400N(sin33))(μ)

Finally, divide Fnet by m, to get the acceleration. (Note that m = weight/g = 820N/g)

a = Fnet/m = Fnet(g/820N)

= ((400N)(cos33) − (820N−400N(sin33))(μ))(g/820N)